FERRELL'S 

ADVAW^OED 

ARITHMETIC 


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Crane  &  Company,  Topeka^  Kansas 


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University  of  California. 


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Accession .9.3.45^ ^^^^ 


Digitized  by  the  Internet  Archive 

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http://www.archive.org/details/ferrellsadvancedOOferrrich 


FERRELL'S 


ADVANCED  AEITHMETIC. 


BY 


J.  A.  FEERELL,  B.  S.,  O.  E., 

Superintendent  of  Sedan  City  Schools,  Sedan,  Kansas,  author  of 

"  Teachers'  and  Students'  Manual  of  Arithmetic," 

and  'Terrell's  Elementary  Arithmetic." 


Cbane  &  Company,  Publishers 

ToPEKA,  Kansas 

1901 


TERRELL'S    ARITHMETICS. 

^kS\  1 

Ferrell's  Teachers'  and  Students'  Manual 
of  Arithmetic. 

165   Pages. 

— : ^'^a  ft.,^ 

Ferrell's  Elementary  Arithmetic. 

217  +  xiii    Pages. 


Ferrell's  Advanced  Arithmetic. 

408  +  xvi    Pages. 


CRANE  &  COMPANY,   PUBLISHERS,  TOPEKA,   KAN. 


(^^^£A^-e^- 


Copyrighted,  1901,  by  Crane  &  Co. 


PREFACE. 


This  book  was  prepared  especially  as  an  advanced  arithmetic 
for  the  public  schools.  It  is  the  companion  book  of  FerrelPs 
Elementary  Arithmetic,  which  contains  the  mathematics  for 
the  third,  fourth,  and  fifth  school  years. 

Briefly,  the  plan  of  the  book  is  as  follows : 

Numbers 


Part  I. 


Part  II. 


Problems 


Number  t 


Part  III. 


Arabic  notation  ( pp.  1-15). 
Processes  (15-78). 

{Equation  (79-92). 
Ratio  (92-99). 
Proportion  (99-105). 
Nature  and  classes  (105-111). 

Methods  of  solution  |  ^-'j^lfi^ll-g^l^^, 

r  Denominate  numbers  (148-179). 
j  Involution  and  evolution  (179-197). 
I  Literal  notation  (197-209). 
'  Formulas  and  their  use  (210-213). 

Mensuration  (213-246). 
[  Problems  J  Percentage  (247-308). 

Mechanics  (308-320). 
i  Miscellaneous  (321-332). 

Positive  and  negative  (333-335). 

Processes  (335-357). 
'  One  base  (357-365). 

Two  bases  (365-386). 

Three  bases  (386-393). 

Quadratics  (393-399). 
,  Progressions  (399-408). 


Numbers 


Problems 


The  author  claims  for  this  book  the  following  special  features 
of  excellence : 

I.  It  is  Complete. — It  comprehends  every  phase  of  the  study 
of  numbers  and  problems,  properly  belonging  to  public-school 

(iii) 

93457 


IV  PREFACE. 

arithmetic.  It  includes  the  fundamental  and  derived  processes 
of  numbers  expressed  by  figures  and  letters  —  integral  and 
fractional,  positive  and  negative.  It  gives  all  the  operations 
of  simple  equations,  and  a  limited  discussion  of  quadratics. 
It  classifies  and  discusses  the  nature  of  all  kinds  of  arith- 
metical problems,  and  illustrates  and  explains  the  various 
methods  of  solution. 

II.  It  is  Scientific. —  If  any  one  thought  more  than  another 
has  inspired  the  author  in  the  preparation  of  this  book,  it  has 
been  this :  There  should  he  system  in  studying  and  teaching 
arithmetic. 

Part  I  gives  the  pupil  (1)  a  thorough  and  practical  discus- 
sion of  the  fundamental  and  derived  processes  (except  involu- 
tion and  evolution)  of  numbers  expressed  by  figures,  and  (2) 
one  complete  method  by  equations  and  one  by  proportion  for 
solving  all  the  simpler  arithmetical  problems.  There  is  not  a 
new  plan  of  reasoning  and  a  new  method  of  solution  for  each 
particular  problem.  Solutions  are  given  complete.  This  part 
contains  the  philosophical  study  of  problems. 

Part  II  (1)  gives  the  pupil  such  a  knowledge  of  expressing 
numbers  by  letters  as  will  enable  him  to  understand  and  inter- 
pret a  formula;  (2)  it  teaches  him  how  to  classify  problems 
into  general  types ;  and  (3)  how  to  develop  and  use  the  formu- 
las for  these  types.     This  part  applies  the  science  of  arithmetic  to 

the  problems  of  practical  Ivi^ivess  life. 

Part  III  gives  the  pupil  (1)  a  knowledge  of  positive  and  nega- 
tive numbers,  (2)  the  method  of  solving  more  difficult  prob- 
lems—  problems  of  one,  two,  three  or  more  unknown  numbers, 
and  (3)  a  limited  knowledge  of  quadratic  equations  and  \)T0- 


PREFACE. 


gressions.     This  part  prepares  the  pupil  for  the  study  of  higher 
mathematics. 

III.  It  IS  Teachable. —  The  teacher  and  the  pupil  are  not 
left  to  guess  at  the  author's  methods  of  presenting  the  various 
subjects.  Every  part  is  explained  and  illustrated,  and  numer- 
ous examples  are  given.  The  pupil  always  has  before  him  a 
model  to  study  and  follow,  and  there  is  no  excuse  for  his  efforts 
being  haphazard  and  aimless.  The  plan  of  the  book  is  adapted 
to  the  developing  mind  both  in  arrangement  and  explanations. 

For  suggestions   on   teaching   this    book,  see   Introduction. 

J.  A.  F. 
Sedan,  Kansas,  August,  1901. 


TABLE    OF    CONTENTS. 


PART  I. 

Page. 

,  Study  of  Numbers 1-78 

A.  Introduction 1-4 

Definitions 1 

B.  Notation  and  numeration  op  integers 4-15 

Definitions 4 

Methods 5 

Figures 6 

Reading  and  writing  numbers  to  one  thousand 6 

Notation  and  numeration  of  integers  in  general 9 

Reading  and  writing  dollars  and  cents 14 

0.  Addition  op  integers 15-18 

Definitions,  signs,  and  principles 15 

Process  of  addition 16 

D.  Subtraction  of  integers 18-20 

Definitions,  signs,  and  principles 18 

Process  of  subtraction 18 

E.  Multiplication  op  integers 21-25 

Definitions,  sign,  and  principles 21 

Process  of  multiplication 23 

F.  Division  op  integers 25-35 

Definitions,  signs,  and  principles 25 

Process  of  short  division 26 

Three  applications  of  division 27 

Different  ways  of  expressing  division 28 

Disposing  of  the  remainder 29 

Long  division 30 

General  review 32 

G.  SiMPLIPYING  numerical  EXPRESSIONS 35-36 

A  term 35 

Process 36 

H.  Derived  operations 37-50 

1.  Factoring 37-44 

Definitions  and  principles 37 

(vii) 


Vlll  CONTENTS. 

Page. 

Factoring  by  inspection 40 

Factoring  by  division 41 

Factoring  expressions  composed  of  two  or  more  terms 42 

Principles  of  multiplication  and  division  relating  to  factor 43 

2.  Cancellation 45-46 

Definition  and  principle 45 

Process 45 

3.  Great  common  divisor 46-48 

Definition  and  principles 46 

Process 47 

4.  Least  common  multiple 48-50 

Definition  and  principles 48 

Process 49 

I.  Common  fractions 51-64 

Definitions  and  principles 51 

Reducing  fractions  to  higher  terms 52 

Reducing  fractions  to  lowest  terms 53 

Reducing  mixed  numbers  to  improper  fractions 54 

Reducing  improper  fractions  to  integers  or  mixed  numbers 55 

Reducing  fractions  to  L.  C.  D 55 

Addition  of  fractions 56 

Subtraction  of  fractions 57 

Multiplication  of  fractions 58 

Division  of  fractions 60 

Greatest  common  divisor  of  common  fractions 62 

Least  common  multiple  of  common  fractions 64 

J.  Decimal  fractions 65-78 

Definitions 65 

Reducing  decimals  to  higher  terms 68 

Reducing  decimals  to  lower  terms 68 

Addition  of  decimals 69 

Subtraction  of  decimals 70 

Multiplication  of  decimals 71 

Division  of  decimals 72 

Reduction  of  common  fractions  to  decimals 74 

Reducing  decimals  to  common  fractions 77 

II.  Study  of  Problems 79-147 

A.  Means  of  expressing  solutions 79-105 

1.  Equations 79-92 

Definitions 79 


CONTENTS.  ,  IX 


Classification  as  to  source v 79 

Transformation 80 

To  turn  an  equation  around 80 

Transposition  of  terms 81 

Simplifying  members 83 

To  multiply  an  equation 83 

To  divide  an  equation 89 

2.  Ratio 92-99 

Definitions 92 

Writing  the  corresponding  abstract  terms  of  a  ratio 93 

Process  of  finding  the  numerical  ratio 97 

3.  Proportion 99-105 

Definitions 99 

Denominations  of  terms 100 

Solving  a  proportion  with  abstract  terms 100 

Solving  a  proportion  with  concrete  terms 102 

B.  Problems  op  one  base  —  nature  and  classification 105-111 

Nature 105 

Stating  the  parts  of  a  problem 106 

Classification 110 

C.  Problems  op  one  basis  —  solution 112-147 

Two  methods  of  solution 112 

1.  Equation  method 112-138 

Arranging  the  parts  of  a  problem 112 

Process  of  solving  problems 116 

Integral  solutions 116 

Fractional  solutions 122 

Reciprocals 125 

Solutions  shortened 127 

Percentage  solutions 131 

2.  Proportion  method 138-147 

Process 138 


PART  II. 

I.  Study  of  Numbers 148-209 

A.  Denominate  numbers 148-168 

Definitions '. 148 

1.  English  system 149-168 

Linear  measures 149 


CONTENTS. 

Page. 

Surface  measures 154 

Solid  measures 156 

Measures  of  capacity 157 

Measures  of  mass 159 

Measures  of  time 162 

Measures  of  value 164 

Review  and  rapid  drill  work 164 

2.  French  system ^69-174 

Linear  measures 169 

Surface  measures 171 

Solid  measures 172 

Measures  of  capacity 172 

Measures  of  mass 174 

Measures  of  value 174 

3.  Compound  denominate  numbers 174-179 

Addition 174 

Subtraction 176 

Multiplication 177 

Division 178 

B.  Involution 179-183 

Definitions 179 

Process 180 

Another  method 181 

C.  Evolution 183-197 

Definition 183 

1.  Square  root 184-190 

First  process 184 

Second  process 184 

Second  process  shortened 187 

2.  Cube  root 190-196 

First  process 190 

Second  process 190 

Second  process  shortened 194 

3.  Roots  of  higher  degree 196-197 

Process 196 

D.  Numbers  expressed  by  letters 197-209 

Explanations 197 

Addition ." 198 

Subtraction 200 

Multiplication 201 


CONTENTS.  ,  XX 

Page. 

Division 202 

Involution , 203 

Evolution 205 

Equations  containing  letters 207 

II.  Study  of  Problems 210-332 

A.  Development  and  use  op  formulas 210-213 

General  numbers  and  general  problems 210 

Solving  the  general  problem  or  type 211 

B.  Mensuration 213-246 

Definitions 213 

1.  Plane  surfaces 216-232 

Parallelograms 216 

Triangles 221 

Other  quadrilaterals 225 

Regular  polygons  and  circles 227 

2.  Solids 232-241 

Polyhedrons 232 

Solids  having  curved  surfaces 237 

Similar  figures 241 

Review 244 

B.  Percentage 247-309 

1.  Percentage  formula 247-251 

Elements  . , 247 

Developing  the  percentage  formula 248 

2.  Percentage  without  time 251-279 

Profit  and  loss 251 

Trade  discount 257 

Commission 261 

Stocks  and  bonds 264 

Taxes 275 

Duties 276 

Insurance 278 

3.  Percentage  with  time 281-308 

Terms  used  in  interest 281 

Time 281 

Simple  interest 284 

Notes 290 

Partial  payments 292 

Annual  interest 294 


Xll  CONTENTS. 

Page. 

Compound  interest 295 

Bank  discount 297 

True  discount 300 

Exchange 302 

0.  Mechanics 308-320 

Force 308 

Work 313 

Activity  or  power 315 

Simple  machines 317 

D.  Miscellaneous 321-332 

Partitive  proportion 321 

Mixture  problems 322 

Equation  of  payments 324 

Partnership 326 

Longitude  and  time 327 


PART  III. 

I.  Study  of  Numbers 333-408 

A.  Integral  numbers 333-351 

Positive  and  negative  numbers 333 

Numbers  expressed  by  means  of  letters 334 

Addition 335 

Subtraction 339 

Multiplication 343 

Division 346 

Factoring 348 

G.  C.  D 350 

L.  C.  M 350 

B.  Fractional  numbers 351-357 

Reduction 351 

Addition 353 

Subtraction 354 

Multiplication 356 

Division 357 

II.  Study  of  Numbers 357-408 

A.  Problems  op  one  basis 357-365 

Solving  equations  of  one  unknown  number 357 

Problems 360 


CONTENTS.  .  Xlll 

Page. 

B.  Problems  op  two  bases 365-386 

Solving  equations  of  two  unknown  numbers 365 

Problems , 373 

C.  Problems  of  three  or  more  bases 386-393 

Solving  equations  of  three  or  more  unknown  numbers 386 

Problems 388 

D.  Problems  containing  quadratic  equations 393-399 

Quadratic  equations 393 

Problems 396 

E.  Progressions 399-408 

Arithmetical  progressions 399 

Falling  bodies 403 

Geometrical  progressions 405 


INTRODUCTION. 

To  the  Teacher: 

I.  Plan. —  Our  best  geometries  first  solve  and  give  full  ex- 
planations of  all  the  fundamental  problems  and  propositions 
of  a  book  (or  part)  of  geometry  for  the  pupil's  study ;  and  then 
give  at  appropriate  places  or  at  the  close  of  the  part  a  list  of 
problems  and  propositions  for  the  pupil  to  solve.  The  logic  of 
this  plan  appears  in  this :  That  the  pupil  is  taught  the  nature 
of  problems  and  correct  methods  of  solving  them  before  he  is 
required  to  solve  them  for  himself. 

This  is  in  general  the  author's  plan  of  teaching  not  only  ge- 
ometry, but  all  the  branches  of  mathematics,  especially  all  of 
the  lower  branches. 

Note.— It  is  bad  pedagogy,  contrary  to  business  judgment,  and  a  great 
waste  of  time,  to  put  a  pupil  at  work  trying  to  solve  problems  before  he 
has  a  knowledge  of  the  nature  of  such  problems  as  he  is  expected  to 
solve,  or  a  knowledge  of  the  plan  by  which  such  problems  are  to  be 
solved. 

Throughout  this  book  the  matter  is  arranged  in  two  divisions 
or  parts:  explanatory  part^  and  exercises.     The  explanatory  part 
consists  of  definitions,  principles,  rules,  explanations,  and  ex- 
amples of  processes  and  solutions.     This  part  is  to  be  learned 
by  the  pupil.     The  exercises  are  to  be  worked  out  by  the  pupil. 

II.  How  TO  Use  the  Explanatory  Part. — Assign  explanatory 
matter  for  study  and  recitation.  (1)  At  the  time  of  assigning  a 
lesson,  go  over  it  with  the  pupil,  and  explain  such  (and  only 
such)  things  as,  in  your  judgment,  your  pupil  will  be  unable 
to  understand  for  himself.  (2)  Require  pupils  to  commit 
definitions  and  principles  to  memory. 

Note. —  Do  not  tolerate  "  parrot"  memorizing.  Before  the  pupil  at- 
tempts to  commit  a  definition,  he  should  find  out  what  it  means.    Half 

(XV) 


XVI  INTRODUCTION. 

of  the  work  of  memorizing  is  done  when  the  pupil  understands  the  mean- 
ing of  the  language  used.  The  author  is  an  advocate  of  memory  work; 
but  let  it  be  intelligent  memory  work,  and  not  ''words,  words,  words." 

(3)  After  reciting  on  principles,  definitions,  and  the  like, 
the  pupil  should  be  sent  to  the  blackboard  without  his  book ; 
and,  after  reaching  the  board,  examples  should  be  assigned  and 
he  be  required  from  his  own  recollection  and  reasoning  to  write 
out  and  thoroughly  explain  the  processes  or  solutions. 

Note. —  It  is  the  least  of  the  author's  intentions  to  permit  the  pupil  to 
pass  over  examples  as  something  to  w^hich  he  may  refer  when  in  trouble  — 
something  to  be  studied  or  neglected  at  his  pleasure.  A  thorough  un- 
derstanding of  the  example  comes  first  in  time  and  importance  in  study- 
ing and  teaching  this  book. 

III.  The  Exercises. — (1)  In  the  number  work,  the  exercises 
are  given  mainly  for  practice.  Accuracy  and  speed  should  be 
the  watchwords  in  number  work.  (2)  In  the  problem  work,  the 
exercises  are  given  to  test  the  pupil's  ability  in  applying  the 
principles,  solutions,  and  formulas  which  he  has  already 
studied;  also  as  a  drill,  better  to  fix  principles,  models,  and 
formulas  in  the  mind. 

IV.  Additional  Exercises. —  Exercise  LVI  has  only  26  prob- 
lems. Nearly  all  the  problems  in  Exercises  CVII  to  CXIV  are 
percentage  problems,  and  the  pupil  should  be  able  to  solve  any 
of  them  by  the  method  given  in  Article  80.  It  is  recom- 
mended that  the  pupil  be  given  a  thorough  knowledge  of  the 
solution  of  percentage  problems  by  the  equation  method. 

V.  Answers. —  Answers  are  not  put  in  this  book.  The  author 
believes  that  the  presence  of  answers,  either  with  the  problems 
or  at  the  close  of  the  book,  fosters  dependency  and  lack  of  self- 
confidence  in  the  pupil.  For  teachers,  who  may  need  answers 
to  save  time  and  labor  in  correcting  and  grading  school  work, 
the  answers  are  printed  in  a  pamphlet,  and  may  be  obtained 
from  Crane  &  Co. 


ADVANCED    ARITHMETIC. 


PART   I. 

I.    STUDY  OF  NUMBERS. 

A.    INTRODUCTION. 

1.  Definitions. — Arithmetic  is  the  introductory 
branch  of  the  science  of  numbers  and  problems. 

Science  is  classified  knowledge. 

A  Number  is  one  or  more  units,  considered  as  forming 
one  quantity,  or  amount.  An  expression  of  one  or  more  units 
is  also  called  a  number.  The  word  number  is  a  term  signifying 
one  or  more  units,  or  an  expression  of  one  or  more  units. 

Note. — A  distinction  should  be  made  between  ^^ number"  as  used  in 
the  sentence,  A  number  of  apples  were  eaten,  where  the  word  number 
means  quantity  or  amount;  and  *' number"  as  used  in  the  sentence,  The 
divisor  is  the  number  on  the  left,  where  the  word  number  means  the  expres- 
sion on  paper. 

A  Unit  is  a  single  thing,  or  one.  Units  are  classed  as  con- 
crete or  abstract,  integral  ov  fractional. 

A  Concrete  Unit  is  one  thing.    As, 

1  book,         1  man,         1  box. 

An  Abstract  Unit  is  simply  one,  a  ratio,  and  not  a  thing. 
As,  J 


XVI  INTRODUCTION. 

of  the  work  of  memorizing  is  done  when  the  pupil  understands  the  mean- 
ing of  the  language  used.  The  author  is  an  advocate  of  memory  work; 
but  let  it  be  intelligent  memory  work,  and  not  ^'words,  words,  words." 

(8)  After  reciting  on  principles,  definitions,  and  the  like, 
the  pupil  should  be  sent  to  the  blackboard  without  his  book ; 
and,  after  reaching  the  board,  examples  should  be  assigned  and 
he  be  required  from  his  own  recollection  and  reasoning  to  write 
out  and  thoroughly  explain  the  processes  or  solutions. 

Note. —  It  is  the  least  of  the  author's  intentions  to  permit  the  pupil  to 
pass  over  examples  as  something  to  which  he  may  refer  when  in  trouble  — 
something  to  be  studied  or  neglected  at  his  pleasure.  A  thorough  un- 
derstanding of  the  example  comes  first  in  time  and  importance  in  study- 
ing and  teaching  this  book. 

III.  The  Exercises. — (1)  In  the  number  work,  the  exercises 
are  given  mainly  for  practice.  Accuracy  and  speed  should  be 
the  watchwords  in  number  work.  (2)  In  the  problem  work,  the 
exercises  are  given  to  test  the  pupil's  ability  in  applying  the 
principles,  solutions,  and  formulas  which  he  has  already 
studied;  also  as  a  drill,  better  to  fix  principles,  models,  and 
formulas  in  the  mind. 

IV.  Additional  Exercises. —  Exercise  LVI  has  only  26  prob- 
lems. Nearly  all  the  problems  in  Exercises  CVII  to  CXIV  are 
percentage  problems,  and  the  pupil  should  be  able  to  solve  any 
of  them  by  the  method  given  in  Article  80.  It  is  recom- 
mended that  the  pupil  be  given  a  thorough  knowledge  of  the 
solution  of  percentage  problems  by  the  equation  method. 

V.  Answers. —  Answers  are  not  put  in  this  book.  The  author 
believes  that  the  presence  of  answers,  either  with  the  problems 
or  at  the  close  of  the  book,  fosters  dependency  and  lack  of  self- 
confidence  in  the  pupil.  For  teachers,  who  may  need  answers 
to  save  time  and  labor  in  correcting  and  grading  school  work, 
the  answers  are  printed  in  a  pamphlet,  and  may  be  obtained 
from  Crane  &  Co. 


ADVANCED    ARITHMETIC 


PART   I. 

I.    STUDY  OF  NUMBERS. 
A.     INTRODUCTION. 

1.  Definitions. — Arithmetic  is  the  introductory 
branch  of  the  science  of  numbers  and  problems. 

Science  is  classified  knowledge. 

A  Number  is  one  or  more  units,  considered  as  forming 
one  quantity,  or  amount.  An  expression  of  one  or  more  units 
is  also  called  a  number.  The  word  number  is  a  term  signifying 
one  or  more  units,  or  an  expression  of  one  or  more  units. 

Note. — A  distinction  should  be  made  between  '^ number"  as  used  in 
the  sentence,  A  number  of  apples  were  eaten,  where  the  word  number 
means  quantity  or  amount;  and  *^ number"  as  used  in  the  sentence,  The 
divisor  is  the  number  on  the  left,  where  the  word  number  means  the  expres- 
sion on  paper. 

A  Unit  is  a  single  thing,  or  one.  Units  are  classed  as  con- 
crete or  abstract,  integral  or  fractional. 

A  Concrete  Unit  is  one  thing.    As, 

1  book,         1  man,         1  box. 

An  Abstract  Unit  is  simply  one,  a  ratio,  and  not  a  thing. 
As,  J 


2  ADVANCED   ARITHMETIC. 

A  unit,  not  considered  as  a  part  of  any  other  unit,  is  an 
Integral  Unit.    As, 

1,         1  book,         1  foot. 

A  unit,  considered  as  a  part  of  some  other  unit,  is  a  Frac- 
tional Unit.     As, 

^  foot,        i  dollar,        -J  pint. 

Note.— The  Hs  1  fifth  of  1  foot.  Then,  the  unit,  ^  is  a  part  of  the 
unit,  1  foot ;  so,  the  i  is  a  part  of  $1  and  the  i  is  a  part  of  1  pint.  The 
unit  of  which  the  fractional  unit  is  a  part  is  called  the  Unit  of  the  Frac- 
tion. 

A  number  composed  of  concrete  units  is  a  Concrete 
Number.     As, 

6  men,         15  dollars,         20  miles. 

A  number  composed  of  abstract  units  is  an  Abstract 
Number.    As, 

6,         15,        20. 

When  a  concrete  number  and  an  abstract  number  have  each 
the  same  number  of  units,  they  are  called  Corresponding  num- 
bers. Every  concrete  number  has  its  corresponding  abstract  num- 
ber. 

Note. — This  fact  will  be  important,  when  you  come  to  solve  problems 
by  proportion  and  by  formulas. 

A  Simple  Number  is  a  number  whose  integral  units 
are  of  the  same  kind  and  size.     As, 

250  bushels,         75  men. 

A  Compound  Number  is  a  number  whose  integral 
units  are  of  the  same  kind  but  of  two  or  more  sizes.     As, 

5  yards  2  feet;         4  bushels  2  pecks  5  quarts. 


ADVANCED   ARITHMETIC.  8 

A  number  composed  of  integral  units  is  an  Integral 
Number,  or  an  Integer.    As, 

12  men,         25. 
Note. — An  integer  may  be  either  concrete  or  abstract. 

A  number  composed  of  fractional  units  is  a  Fractional 
Number,  or  a  Fraction.    As, 

f  dollar,        t5_. 
Note. — A  fractional  number  may  be  either  concrete  or  abstract. 

A  number  composed  partly  of  integral  units  and  partly  of 
fractional  units  is  a  Mixed  Number.    As, 

5i  dollars,         Of. 

Note. — A  mixed  number  may  be  either  concrete  or  abstract. 

A  Problem  is  a  question  proposed  for  solution. 

Note. — A  problem  is  not  always  stated  in  the  form  of  a  question.  For 
example, 

Find  the  cost  of  10  books,  at  SOf  each. 

But  such  problems  have  all  the  essentials  of  questions,  and  may  be 
easily  so  stated.    Thus, 

What  is  the  cost  of  10  hooks,  at  SOf  each  f 

EXERCISE  I. 

1.  What  is  arithmetic  ? 

2.  Are  there  other  branches  of  the  science  of  number  ? 

3.  What  is  meant  by  "  introductory  branch  "  ? 
4..  What  is  science  f 

5.  What  then  is  the  science  of  number  ? 

6.  What  is  a  number  ? 

7.  What  is  quantity  f 

8.  What  is  a  unit  f 


4  ADVANCED    ARITHMETIC. 

9.  In  the  sentence,  The  number  on  the  right  is  the  quotient, 
does  the  word  number  signify  one  or  more  units  or  does  it  sig- 
nify the  expression  that  is  written  ? 

10.  In  the  sentence,  I  do  not  know  the  number  of  marbles 
that  were  lost,  is  the  word  number  and  the  real  number  of  mar- 
bles the  same  thing  ?     What  is  the  difference,  if  any  ? 

11.  In  the  sentence,  There  are  7  books  on  the  table,  is  the  ex- 
pression, "7  books,"  and  the  real  number  of  books  the  same 
thing  ?    What  is  the  difference,  if  any  ? 

12.  Number  is  sometimes  defined  as  a  collection  of  units. 
Do  the  units  have  to  be  collected,  or  may  they  be  considered 
together  as  forming  one  quantity  without  being  in  reality  col- 
lected ?     Illustrate  your  answer  with  an  example. 

13.  Give  the  classification  of  units. 

H.  What  is  a  concrete  unit  ?    Give  an  example. 

15.  What  is  an  abstract  unit  ?    Give  an  example. 

16.  What  is  an  integral  unit  ?     Give  an  example. 

17.  What  is  2i  fractional  unit  f     Give  an  example. 

18.  What  is  a  concrete  number  ?     Give  an  example. 

19.  What  is  an  abstract  number  ?     Give  an  example. 

20.  What  are  corresponding  numbers  ?    Give  an  example. 

21.  What  is  an  integer?    Give  an  example. 

22.  What  is  a  fraction  f    Give  an  example. 

23.  What  is  the  unit  of  a  fraction  ?     Give  an  example. 
2J/..  What  is  a  mixed  number?    Give  an  example. 

25.  What  is  a  problem  ? 

26.  Is  a  problem  always  in  the  form  of  a  question  ?  Illus- 
trate your  answer  by  examples. 

27.  Are  all  questions  problems  ?  Illustrate  your  answer  by 
examples. 

B.    NOTATION  AND  NUMERATION  OF  INTEGERS. 

2.  Definitions. — ]N"otation  is  the  process  of  express- 
ing numbers  by  means  of  words,  letters,  or  figures. 


ADVANCED   ARITHMETIC.  5 

Numeration  is  the  process  of  reading  numbers,  expressed 
in  words,  letters,  or  figures. 

Note. — A  pupil  that  can  correctly  write  numbers  can  read  them,  for 
the  same  knowledge  of  the  method,  of  use  of  terms  and  characters,  is  re- 
quired in  both.    Hence,  both  processes  are  treated  together  in  this  book. 

3.  Metliods. — Five  methods  of  expressing  numbers  are 
in  common  use : 

(1)  By  means  of  words  that  express  the  number  of  units 
considered.     As, 

three,        nine,        twenty-five. 

(2)  By  means  of  letters  that  express  the  number  of  units 
considered.     As, 

III,        IX,        XXV. 

Note. — Neither  of  the  above  methods  is  used  in  arithmetical  compu- 
tations.   Their  study  belongs  to  the  reader  rather  than  to  the  arithmetic. 

(3)  By  means  of  words  that  do  not  express  the  number  of 
units  considered.     As, 

cost  price,        John's  age,        B's  money. 

Note. — This  method  of  expressing  number  is  used  in  the  solution  of 
arithmetical  problems.  It  is  self-explanatory,  and  does  not  require  spe- 
cial study  at  this  time. 

(4)  By  means  of  letters  that  do  not  express  the  number  of 
units  considered.     As, 

X,         2  ay,         m*. 

Note. — This  method  is  usually  called  algebraic  notation.  It  is  studied 
and  used  in  Parts  II  and  III. 

(5)  By  means  of  figures  that  express  the  number  of  units 
considered.     As,  o  o  ok 

Note.— This  method  as  herein  used  is  called  the  Arabic  method.  It 
is  the  common  method  of  expressing  known  numbers  in  arithmetical 
computations. 


6  ADVANCED   ARITHMETIC. 

4.  Figures. — The  Arabic  method  of  notation  employs  ten 
characters,  called  Figures  or  Digits.  Their  forms  and 
names  are  as  follows : 


0, 

1, 

2, 

s, 

4, 

5, 

6, 

7,       8,       9. 

zero, 

one, 

two. 

three, 

four. 

five, 

six, 

seven,  eight,  nine. 

Note. — The  0  is  also  called  naught.  Some  authors  claim  that  there 
are  only  nine  digits,  and  that  0,  representing  no  value,  is  not  a  digit. 

5.  Reading  and  Writing  Integers  to  One 
Tliousand. — In  reading  and  writing  numbers  by  the  Arabic 
method — 

( 1 )  The  integers  up  to  ten  are  named  and  written  with  fig- 
ures as  follows : 

one,      1  five,      5 

two,      2  six,        6 

tliree,  3  seven,  7 

four,    4  eiglit,   8 

nine,  9 

(2)  In  integers  of  more  than  nine  units,  the  units  are  grouped 
as  far  as  possible  into  groups  of  ten  each.  Groups  of  tens  up 
to  ten  tens  are  named  and  written  with  figures  as  follows : 

1  ten,   ten,           10  5  tens,  fifty,          50 

2  tens,  twenty,  20  6  tens,  sixty,  60 
8  tens,  tliirty,  30  7  tens,  seventy,  TO 
4  tens,  forty,       40  8  tens,  eiglity,     80 

9  tens,  ninety,  90 

(3)  In  integers  of  10  tens  or  more,  the  groups  of  ten  each 
are  grouped  as  far  as  possible  into  larger  groups  of  10  tens 
each.     One  of  these  larger  groups  is  called  a  hundred.     Hun- 


ADVANCED   ARITHMETIC. 


dreds  up  to  10  hundreds  are  named  and  written  with  figures  as 
follows : 

one  liundred,      100  five  liundred,      500 

two  liundred,     200  six  liundred,        600 

three  hundred,  300  seven  hundred,  TOO 

four  hundred,    400  eight  hundred,  800 

nine  hundred,  900 

(4)  Integers  between  10  and  20  are  named  and  written  with 


igures  as  follows : 

1  ten  1  unit; 

eleven ; 

11 

1  ten  2  units 

twelve ; 

12 

1  ten  3  units 

thirteen  ; 

13 

1  ten  4  units 

fourteen ; 

14 

1  ten  5  units 

fifteen  ; 

15 

1  ten  6  units 

;   sixteen ; 

16 

1  ten  7  units 

seventeen ; 

17 

1  ten  8  units 

;   eighteen  ; 

18 

1  ten  9  units 

nineteen ; 

19 

(5)  Any  integer  between  20  and  100  is  written  with  figures 
by  placing  the  number  of  tens  in  the  second  place  from  the 
right  or  tens^  place  and  the  number  of  units  in  the  right-hand 
j)lace  or  units^  place.  If  there  are  no  units,  place  a  0  in  units' 
place.  These  numbers  are  read  by  naming  the  tens  and  units 
in  succession.     Thus: 

21,    twenty-one  47,    forty-seven 

78,    seventy-eight  86,    eighty-six 

55,    fifty-five  34,    thirty-four 

92,    ninety-two  63,    sixty-three 

99,    ninety-nine 

(6)  Any  integer  between  1  hundred  and  10  hundred  is  writ- 
ten by  placing  the  number  of  hundreds  in  the  third  place  from 


O  ADVANCED    ARITHMETIC. 

the  right,  or  hundreds''  j)lace  ;  the  number  of  tens  in  tens'  place ; 
and  the  number  of  units  in  units'  place.  Fill  vacant  places 
with  O's.  These  numbers  are  read  by  naming  the  hundreds, 
tens,  and  units  in  succession.     Thus: 

125,  one  liundred  twenty-five 

790,  seven  liundred  ninety 

467,  four  liundred  sixty-seven 

806,  eiglit  liundred  six 

512,  iive  liundred  twelve 

272,  tTvo  liundred  seventy-two 

638,  six  hundred  thirty-eiglit 

301,  tliree  liundred  one 

999,  nine  hundred  ninety-nine 

Note. —  Do  not  use  "  and  "  in  reading  these  numbers. 

Since,  1  liundred  =  10  tens,  and 

1  ten  =  10  units, 

a  figure  in  hundreds'  place  represents  10  times  as  many  as  the 
same  figure  in  tens'  place;  a  figure  in  tens'  place  represents  10 
times  as  many  as  the  same  figure  in  units'  place. 


EXERCISE  II. 

Read  rapidly : 

1. 

2. 

3. 

4. 

5. 

78 

34 

800 

065* 

259 

19 

60 

139 

815 

734 

25 

71 

240 

725 

813 

82 

88 

327 

899 

004* 

96 

07* 

444 

868 

117 

57 

56 

516 

992 

629 

43 

99 

607 

408 

999 

*  In  integers,  a  0  standing  on  the  left  of  other  figures  has  no  effect  upon 
the  number.    07  is  7.    065  is  65.     004  is  4. 


ADVANCED    ARITHMETIC. 


Write  : 

6. 

7. 

8. 

9. 

twenty-one 

fifty-four 

thirty-nine 

sixty-nine 

seventy 

ninety 

eighty-one 

thirty-seven 

eighty-nine 

seventy-two 

thirteen 

ninety-three 

thirty-eight 

nineteen 

seventy-three 

forty-five 

forty-six 

sixty-six 

ninety-eight 

eighty-four 

10. 
Four  hundred  ninety-four 
Two  hundred  nineteen 
Five  hundred  fifty-five 
Seven  hundred  seventy-nine 
Eight  hundred  six 
Nine  hundred  sixty-nine 
Eight  hundred  sixty 
Seven  hundred  ten 
Six  hundred  sixty-six 
Four  hundred 


11, 

Eight  hundred  seventy- two 
Three  hundred  thirty-three 
One  hundred  fourteen 
Nine  hundred  three 
Eight  hundred  seventy-six 
Six  hundred  twenty-four 
Five  hundred  nineteen 
Seven  hundred  ninety-four 
Two  hundred  forty-eight 
Nine  hundred  ninety-nine 


6.  Notation  and  Numeration  of  Integ-ers  in 
Greneral. —  By  the  Arabic  method,  integers  larger  than  nine 
are  represented  by  figures  placed  side  by  side.     Thus: 

274139568 


The  8  represents  I's,  or  units  of  the  first  order  ; 
The  6  represents  lO's,  or  units  of  the  second  order; 
The  5,  units  of  the  tliird  order  ; 
The  9,  units  of  the  fourtli  order  ; 

and  so  on. 

The  order  of  any  figure  or  unit  is  the  number  of  its  place, 
counting  from  the  right. 


>i 


10  ADVANCED   ARITHMETIC. 

This  method  of  representing  integers  is  based  upon  the  fol- 
lowing law : 

Law. — Ten  units  of  any  one  order  make  one  unit  of  the  next 
higher  order. 

In  reading  and  writing  integers,  the  figures  are  considered, 
as  far  as  possible,  iij  groups  of  three  figures  each,  commencing 
at  the  right  hand.  These  groups  are  called  Periods.  The 
first  twelve  periods,  counting  from  the  right,  are  named  as 

follows : 

units  quintillions 

tliousands  sextillions 

millions  septillions 

billions  octillions 

trillions  nonillions    . 

quadrillions  decillions 

Note. — If  the  word,  period,  follows  the  name,  the  possessive  sign  should 
be  used.     Either  period  of  thousands  or  thousands'  period  is  correct. 

Let  it  be  required  to  read  the  number, 
346524207985325. 
Separating  it  into  periods,  and  naming  the  periods,  we  have — 

trillions^    billions,    millions,    thousands,    units. 

346,      524,      207,       985,      325. 

How  many  units  ?     Ans.,  325. 
How  many  thousand  ?     Ans.,  985. 
How  many  million  ?     Ans.,  207. 
How  many  billion  ?     Ans.,  524. 
How  many  trillion  ?     Ans.,  346. 

Then  read  the  number,  beginning  with  the  left-hand  period. 
Thus,  346  trillion,  524  billion,  207  million,  985  thousand,  325  (units). 

Note. —  In  reading  numbers,  drop  the  word  units,  and  say  hundred, 
thousand,  million,  &c,,  not  hundreds,  thousands,  millions,  &c.  Never  use 
"  and  "  in  reading  integers. 

The  figures  in  each  period  are  read  by  methods  explained  in  Article 
5  ;  then  the  name  of  the  period  is  added. 


ADVANCED   ARITHMETIC.  11 

Let  it  be  required  to  write  seventy-five  hillion,  nine  hundred 
forty-eight  million,  five  hundred  one  thousand,  three  hundred  sixty- 
two. 

How  many  billion  ?     Arts.,  75. 

How  many  million  ?     Ans.,  948. 

How  many  thousand  ?     Ans.,  501. 

How  many  units  ?     Ans.y  362. 

Then  write  the  number.     Thus,  75,948,501,362. 

Note.— When  it  is  desired  to  separate  the  periods  of  a  number,  com- 
mas are  used ;  as  in  the  last  number  above. 

Since  a  period  has  three  places,  the  place  on  the  right  hand 
is  units  of  the  period,  the  second  place  is  tens  and  the  third 
place  is  hundreds.  To  illustrate,  the  first  place  in  thousands  is 
units  of  thousands,  or  simply  thousands;  the  second  place,  ten- 
thousands  ;  the  third  place,  hundred-thousands.  The  first  place 
in  millions  is  millions;  the  second,  ten-millions;  the  third, 
hundred-millions;   and  so  on. 

The  method  of  grouping  three  figures  into  a  period,  as  ex- 
plained above,  is  the  method  used  by  the  people  of  the  United 
States  and  France,  and  is  called  the  Frencli  Method. 
The  English  people  and  people  of  many  other  European  na- 
tions consider  a  period  as  made  up  of  six  figures.  This  method 
of  grouping  the  figures  is  called  the  Englisli  Metliod. 
The  first  six  periods  according  to  this  method  are  named  as 
follows : 

units  trillions 

millions  quadrillions 

billions  cxuintillions 

Let  it  be  required  to  read  346524207985825  by  the  English 
method. 


12  ADVANCED   ARITHMETIC, 

Separating  the  number  into  periods,  we  have  — 

billions,        millions,  units. 

846,      524207,      985325 

How  many  units  ?     Ans.,  985325. 
How  many  million  ?     Ans.,  524207, 
How  many  billion  ?     Ans.,  346. 

Then  read  the  number,  beginning  with  the  left-hand  period. 
Thus,  346  billion,  524207  million,  985325  (units). 

Note. — The  figures  in  each  period  are  read  by  the  methods  already 
explained,  then  the  name  of  the  period  is  added. 


EXAMPIiES. 

1.  Read  3462465075. 

Process  :  3,462,465,075;  3  bilHon,  462  million,  465  thousand,  75. 

Note.— Use  the  French  method  unless  the  English  method  is  called 
for  or  indicated.  Commas  are  not  commonly  used  to  separate  the  peri- 
ods, unless  there  are  three  or  more  periods.  In  smaller  numbers,  the 
periods  are  located  mentally. 

2.  Read  672000024941. 

Process  :  672,000,024,941 ;  672  billion,  24  thousand,  941. 

S.  Read  873002403400540  (  English  method  ). 

Process :  873,002403,400540;  873  billion,  2403  million,  400540. 

i.  Write  894  billion,  1  thousand,  127. 

Written :  894,000,001,127. 

Note. — There  are  no  millions ;  fill  the  orders  with  O's.  There  are  no 
hundreds  of  thousands  or  tens  of  thousands ;  fill  these  orders  with  O's. 

5.  Write  1753  billion,  201400  million,  843004. 
Written :  1753,201400,843004. 


ADVANCED   ARITHMETIC.  IB 

6.  Write  5  units  of  the  6th  order,  7  units  of  the  3d  order,  1 
unit  of  the  2d  order,  8  units  of  the  1st  order. 
Written:  500718. 
Note.— Fill  the  5th  and  4th  orders  with  O's. 

EXERCISE  III. 

i.  What  is  the  law  upon  which  the  Arabic  method  of  read- 
ing and  writing  numbers  is  based  ? 

2.  What  is  the  order  of  a  figure  ?     How  are  orders  numbered  ? 

S.  How  many  figures  in  a  period  according  to  the  French 
method  ?  Where  do  you  commence  to  point  off  a  number  into 
periods  ? 

4-.  How  many  figures  in  a  period  according  to  the  English 
method  ? 

5.  Name  in  order  twelve  periods  of  the  French  method. 

6.  Name  in  order  six  periods  of  the  English  method. 

7.  Do  the  French  and  English  methods  both  conform  to  the 
law  of  the  Arabic  method  ? 

Read  by  the  French  method  : 

8.  43250  13,   78604501257943 

9.  678426  U,    10030040250641 

10.  9407081  15.   83000034562135 

11.  5675694073  i^.'  786429536742 

12.  889970708432  17.   182103243576842 

Read  by  the  English  method : 

18.  765302456901327. 

19.  1234567878785821. 

20.  986485200840561256. 

Write : 

21.  55  thousand,  421. 

22.  704  million,  78  thousand,  25. 


14  ADVANCED   ARITHMETIC. 

23.  846  billion,  721  million,  1  thousand. 
24..  Twenty-five  thousand  seven  hundred  one. 

25.  Seventy-six  billion,  eight  hundred  one  million,  sixteen 
thousand  three  hundred  seven. 

26.  Seven  units  of  the  5th  order,  nine  units  of  the  4th  order, 
six  units  of  the  8d  order,  one  unit  of  the  2d  order,  8  units  of  the 
1st  order. 

27.  8  units  of  the  7th  order,  2  units  of  the  5th  order,  5  units 
of  the  4th  order,  9  units  of  the  1st  order. 

28.  125  quadrillion,  125  trillion,  125  million,  125. 

29.  Five  hundred  sixty-six  thousand  two  hundred  eighty-one 
million,  eighty-nine  thousand  twenty-one. 

30.  675421  billion,  604001  million,  845762. 

7.  Reading'  and  Writing  Dollars  and  Cents.— 

The  dollar-mark,  $,  is  always  placed  before  the  number  with 
which  it  is  used.     Thus, 

,  $5,  five  dollars. 
$908,  nine  hundred  eight  dollars. 

Cents  may  be  expressed  with  dollars  or  with  a  dollar-mark. 
When  so  expressed,  tivo  places  are  always  used  for  cents. 

100  cents  =  1  dollar;  then, 

any  number  of  cents  less  than  a  dollar  may  be  written  with  two 
figures,  of  which  the  right-hand  figure  is  for  the  units  of  cents, 
and  the  left-hand  figure  is  for  the  tens  of  cents  (dimes).  A 
period  (.)  is  always  placed  between  dollars  and  cents  and  the 
word  and  is  used  there  in  reading.     Thus, 

$5.10,  5  dollars  and  10  cents; 
$12.06,  12  dollars  and  6  cents; 
$.05,  5  cents; 
$807.50,  807  dollars  and  fifty  cents. 


ADVANCED    ARITHMETIC.  15 

Cents  may  be  written  by  using  the  cent-mark,  /.     Thus, 
3/,  3  cents ;         20/,  20  cents ;         85/,  85  cents. 


EXERCISE  IV. 

Read  : 

1. 

2. 

S. 

4. 

$7 

$10.10 

$.40 

25/ 

$59 

$72.55 

$.75 

5/ 

$408 

$246.05 

$.03 

70/ 

$3906 

$109.13 

$.99 

99/ 

5,  Write  with  $  :  Two  dollars  and  forty  cents ;  seventeen  dol- 
lars and  seventy-five  cents ;  nine  thousand  five  hundred  sixty- 
three  dollars  and  twenty-five  cents ;  two  cents ;  fifty  cents. 

6.  Write  with  / ;  Twenty-five  cents ;  nine  cents ;  eighteen 
cents;  one  hundred  twenty-five  cents ;  ninety-five  cents. 

C.    ADDITION  OF  INTEGERS. 

8.  Definitions,  Signs,  and  Principles. — Addi- 
tion is  the  process  of  uniting  two  or  more  numbers  into  one. 
The  numbers  to  be  added  are  called  Addends.  The  result 
of  addition  is  called  the  Sum.  or  Amount. 

The  Sign  of  Addition,  +,  is  read  plus. 

The  Sign  of  Equality,  =,  is  read  equals. 

17+9  =  26 

is  read  17  plus  9  equals  26,  and  means  that  9  added  to  17  gives  a  result 
of  26.    17  and  9  are  addends,  and  26  is  the  sum. 

In  the  fundamental  processes  of  mathematics,  certain  self- 
evident  principles  must  be  observed.  In  addition  they  are  as 
follows : 

Principles:  1.   Only  similar  numbers  can  be  added, 

2.  The  sum  contains  all  the  units  of  all  addends. 

3.  The  sum  and  the  addends  must  be  similar  numbers. 


16  ADVANCED    ARITHMETIC. 

If..  The  sum  is  the  same,  regardless  of  the  order  in  which  the  num- 
bers are  added. 

Note. —  Similar  numbers,  as  here  used,  are  those  whose  integral  units 
are  of  the  same  kind  and  size. 

9.  Process  of  Addition. — Write  the  addends  so  that 
figures  of  the  same  order  will  stand  in  a  column.  Always  be- 
gin with  units  to  add. 

Ten  units  of  one  order  make  one  of  the  next  higher  order.  Then, 
in  adding  numbers  of  more  than  one  order,  if  the  sum  of  all 
the  units  of  any  order  is  ten  or  more,  the  tens  of  that  order 
are  considered  as  units  of  the  next  higher  order,  and  are  so 
added. 

Add  576,  842,  75,  981,  203. 

Process.  Explanation  :  (1)  The  sum  of  the  units  of  the  first  col- 

576  umn  (the  one  on  the  right)  is  17,  or  1  unit  of  the  2d  order 

342  and  7  units  of  the  first  order.    Write  the  7  below,  and 

po?  carry  the  1  to  the  next  column. 

203  (2)  The  sum  of  the  units  of  the  2d  column,  including 

2177  result        *^^  ^  carried,  is  27,  or  2  units  of  the  3d  order  and  7  units 

of  the  2d  order.     Write  the  7  below  and  carry  the  2  to  the 

next  column. 

(3)  The  sum  of  the  units  of  the  3d  column,  including  the  2  carried,  is 

21,  or  2  units  of  the  4th  order  and  1  unit  of  the  3d  order.     Write  the  1 

and  the  2  below  in  their  proper  orders. 

Tests  of  Accuracy:  (1)  Revieiv  the  work,  or  (2)  add  the  col- 
umns in  reverse  order. 

Note. — Various  other  tests  of  accuracy  in  addition  are  suggested  by 
different  authors.  Such  as,  casting  out  the  9's,  or  by  separating  the 
addends  into  two  or  more  groups,  adding  each  group,  then  adding 
their  sums.  Such  methods  are  not  resorted  to  once  in  a  thousand 
times  to  test  the  accuracy  of  addition. 


ADVANCED   AEITHMETIC, 


17 


EXERCISE 

V. 

Copy  and  find  sums  : 

1. 

2, 

3. 

^. 

5, 

6. 

346 

76d 

555 

276 

821 

482 

425 

675   - 

462 

549 

874 

888 

879 

578 

888 

695 

999 

476 

276 

989 

951 

598 

846 

827 

942 

876 

877 

846 

594 

848 

379 

768 

894 

589 

666 

555 

566 

898 

856 

444 

389 

776 

7. 

8. 

9. 

10. 

11, 

12. 

942 

872 

584 

829 

938 

456 

224 

884 

882 

559 

986 

472 

837 

724 

486 

629 

729 

829 

246 

654 

785 

478 

259 

464 

461 

542 

682 

545 

454 

928 

729 

421 

428 

882 

793 

846 

288 

899 

987 

765 

455 

999 

13. 

U^ 

15. 

16. 

17. 

12467 

99999 

95487 

57684 

88579 

86492 

86897 

58987 

66666 

54865 

94876 

58987 

89987 

57868 

47986 

87594 

68859 

66666 

47984 

55555 

59187 

45958 

97869 

88946 

84484 

67895 

78986 

77777 

98979 

98897 

78957 

54852 

12345 

78787 

48864 

98798 

77958 

99999 

57978 

87758 

78978 

68879 

98765 

86868 

69589 

67594 

59987 

88888 

69699 

57868 

98979 

68827 

48767 

87879 

48573 

64888 

77989 

56789 

87654 

48452 

18 


ADVANCED   ARITHMETIC. 


18. 

123456789 
547695487 
865796543 

759876987 
666666666 
989898989 
465798788 
543543543 
999988888 
789789789 


19. 

876543219 

34768492 

99999 

7586493 

59876538 

789437598 

888888888 

958764356 

579867958 

576576576 


24678 

579642 

8978687 

59875879 

4769876543 

989898989 

65943756 

987689 

8976576438 

9578495789 


D.     SUBTRACTION  OF  INTEGERS. 

10.  Definitions,  Sign,  and  Princii)les. —  Sub- 
traction is  the  process  of  taking  one  number  from  another. 
The  number  taken  is  called  the  Subtraliend.  The  number 
from  which  the  subtrahend  is  taken  is  called  the  Minuend, 
The  result  of  subtraction  is  called  the  Remainder  or  Dif- 
ference. 

The  Sign  of  Subtraction,  — ,  is  read  minus. 

Thus,  17-9  =  8 

is  read  17  minus  9  equals  8,  and  means  that  9  taken  from  17  leaves  8. 
17  is  the  minuend  ;  9,  the  subtrahend  ;  and  8j  the  remainder. 

Principles:  1.  The  minuend,  subtrahend,  and  remainder  must 
be  similar  numbers. 

2.  The  remainder  equals  the  minuend  minus  the  subtrahend. 

3.  The  subtrahend  equals  the  minuend  minus  the  remainder. 

}f.   The  minuend  equals  the  sum  of  the  subtrahend  and  remainder. 

11.  Process  of  Subtraction. —  Place  the  subtrahend 
under  the  minuend,  units  under  units,  tens  under  tens,  etc. 
Always  begin  with  units  to  subtract. 

One  of  any  order  (except  the  first  order)  makes  ten  of  the  next 


ADVANCED   ARITHMETIC »  19 

lower  order.  When  a  figure  of  the  minuend  represents  fewer 
units  than  the  corresponding  figure  of  the  subtrahend,  one 
unit  is  taken  ("borrowed"-)  from  the  next  higher  order  of 
the  minuend  and  considered  as  ten  of  the  lower  order. 


EXA1CPI.ES. 

1.  From  43  take  26. 

Process.  Explanation  complete  :  ( 1 )  "We  cannot  take  6  units  from  3 
43  units ;  but  we  can  take  one  ten  of  the  4  tens,  and  add  it  to  the 
26  3  units.  This  will  make  13  units.  Then,  13  units -6  units  =  7 
]7         units.    Write  the  7  below.    (2)   One  ten  has  been  taken  from 

the  4  tens,  which  leaves  but  3  tens  in  the  minuend.    3  tens -2 

tens  leaves  1  ten.    Write  the  1  below. 

2.  From  704  take  875. 

Process.  Explanation  complete:  (1)  We  cannot  take  5  units  from  4 
704  units,  and  there  are  no  tens  in  the  minuend ;  then  we  must 
375  take  1  hundred  from  the  7  hundred :  1  hundred  =  10  tens. 
329         Now,  we  can  take  1  of  the  10  tens  and  add  it  to  the  4  units. 

1  ten +4  units  =  14  units.     14  units -5  units  =  9  units.     Write 

the  D  below. 

(2)  Of  tho  10  tens  we  took  from  hundreds'  place,  we  have  used  1  ten 
and  have  9  tens  left.    9  tens  -  7  tens  =  2  tens.    Write  the  2  below. 

(3)  We  have  used  1  hundred  of  the  7  hundred ;  then  we  have  only  6 
hundred  left.    6  hundred  -  3  hundred  =  3  hundred.    Write  the  3  below. 

3.  From  8801  take  5674. 

Process.        Explanation  shortened :  (1)  Looking  upon  1,  think  11  (why  ? ). 
8801         11  -  4  =  7.    Write  the  7  below. 

5674  (2)  Looking  upon  the  0,  think  10,  then  9  (why?).    9-7  =  2. 

3127         Write  the  2  below. 

(3)  Looking  upon  the  ^,  think  7  (why?).    7-6  =  1.    Write 
the  1  below. 

(4)  8-5  =  3.    Write  the  .?  below. 

Tests  of  Accuracy:  (1)  Review  the  work;  or  (2)  the  minu- 
end minus  the  remainder  should  equal  the  subtrahend;  or  (3)  the 
sum  of  the  subtrahend  and  remainder  should  equal  the  minuend. 


20 


ADVANCED   ARITHMETIC. 


EXERCISE 

VI. 

Copy  and  find  remainders : 

1. 

^. 

3. 

-^. 

5. 

764 

591 

932 

846 

821 

178 

486 

763 

487 

295 

6, 

7. 

8. 

9. 

10. 

887 

700 

500 

800 

400 

639 

583 

479 

865 

128 

11, 

12. 

IS. 

U^ 

15. 

8426 

7363 

8342 

1264 

1848 

7544 

5486      ' 

486 

947 

796 

16. 

i7. 

18. 

19. 

20. 

1920 

1980 

2004 

2245 

6472 

749 

806 

1759 

1892 

2808 

21. 

22. 

23. 

2J^. 

25. 

7038 

8388 

1675 

2008 

3004 

5984 

2845 

597 

1009 

2106 

Find  remainders : 

26.  94675- 

-67897 

37. 

345678-95876 

27.  58406- 

-57643 

38. 

193846-87909 

28.  70698- 

-67899 

39. 

150000-74567 

29.  54868- 

-48794 

1,0. 

183456-89999 

30.  60054- 

-54876 

Ul. 

100000-65482 

31.  92845- 

-56789 

1,2. 

240000-186426 

,^^.  80125- 

-48056 

U3. 

170984-145096 

33.  72010- 

-35476 

U^ 

333888-278747 

,5^.  82904- 

-68926 

J,5. 

811221-456789 

55.  92500- 

-47882 

U6. 

830405-640500 

36.  76543- 

-34567 

J^7. 

900002-712345 

ADVANCED    ARITHMETIC.  21 

E.     MULTIPLICATION  OF  INTEGERS. 

12.  Definition,  Sign,  and  Principles.— Multi- 
plication is  a  process,  shorter  than  addition,  for  finding  the 
sum  when  one  number  is  to  be  used  as  an  addend  several  times. 
The  number  to  be  used  as  the  addend  is  called  the  Multipli- 
cand. The  number  showing  how  many  times  the  multipli- 
cand is  to  be  used  is  called  the  Multiplier.'  The  result 
of  multiplication  is  called  the  Product.  The  expression, 
^^  multiplied  by,''^  used  to  indicate  multij^lication,  means  that 
the  number  placed  before  it  is  to  be  used  as  the  multiplicand 
and  the  number  placed  after  it  is  to  be  used  as  the  multiplier. 

The  expression,  ^^  multiplied  together,^ ^  is  u^ed  to  indicate  the 
multiplication  of  one  number  by  another  without  designating 
which  number  is  to  be  used  as  the  multiplier  or  multiplicand. 

The  Sign  of  Multiplication,  X,  is  read  times* 
*'iX"is  read  one  time,  or  once;   ";^X,"  two  times,  or  twice. 

5x7  =  35 

is  read  5  times  7  equals  35,  and  means  that  the  sum  obtained  by  using  7 
as  an  addend  5  times  is  35.  The  5  is  the  multiplier  ;  the  7,  the  multipli- 
cand ;  and  the  35,  the  product. 

Principles  :  1.  The  multiplicand  may  be  either  concrete  or  ab- 
stract. 

2.   The  multiplier  must  be  abstract. 

8.   The  product  must  be  similar  to  the  multiplicand. 

If.  When  the  two  numbers  to  be  multiplied  together  are  both  ab- 
stract, the  product  is  the  same,  whichever  number  is  taken  as  the 
multiplier. 

Note. —  Before  proceeding  with  multiplication,  the  pupil  should  have 
a  thorough  knowledge  of  the  multiplication  table. 

*Many  authors  also  read  the  sign,  X,  multiplied  by.  In  that  reading,  the  multiplicand 
must  come  before  and  the  multiplier  after  the  sign. 

For  reasons  which  It  would  not  be  profitable  to  discuss  here,  the  author  prefers  to  read  the 
sign  times^  and  adheres  to  that  reading  throughout  this  book. 


22 


ADVANCED    ARITHMETIC. 


MULTIPLICATION   TABLE. 

IX     0=    0 

2X    0=    0 

8X    0=   0 

4X    0=   0 

ix    1=    1 

2X    1=   2 

8X    1=   8 

4X    1=   4 

IX    2=   2 

2X    2=   4 

8X    2=   6 

4X    2=   8 

IX    8=    3 

2X    8=   6 

8X    8=   9 

4X    8  =  12 

IX    4=   4 

2X    4=   8 

8X   4=12 

4X   4=16 

IX    5=    5 

2X    5  =  10 

8X    5  =  15 

4X    5  =  20 

IX    6=    6 

2X    6=12 

8X    6  =  18 

4X    6  =  24 

IX    7=   7 

2X    7=14 

8X    7  =  21 

4X    7  =  28 

IX    8=   8 

2X    8  =  16 

8X    8  =  24 

4X    8  =  82 

IX    9=    9 

2X    9=18 

8X   9  =  27 

4X    9  =  86 

IX  10=10 

2X10  =  20 

8X10  =  80 

4x10  =  40 

1X11  =  11 

2X11=22 

8X11=88 

4x11=44 

1X12=12 
5X    0=   0 

2  X  12  =  24 

8x12  =  86 

4X12  =  48 
8X   0=   0 

6X    0=   0 

7X   0=   0 

5X    1-   5 

6X    1=   6 

7x    1=   7 

8X    1=    8 

5X    2  =  10 

6X    2  =  12 

7x   2  =  14 

8X    2=16 

5X    8  =  15 

6X    8  =  18 

7x    8  =  21 

8X    8  =  24 

5X   4  =  20 

6X    4  =  24 

7X   4  =  28 

8X   4  =  82 

5X   5  =  25 

6X    5  =  80 

7X   5  =  85 

8X   5  =  40 

5X    6  =  80 

6X   6  =  86 

7X    6  =  42 

8X    6  =  48 

5x    7  =  85 

6X    7  =  42 

7x    7  =  49 

8X    7  =  56 

5x   8  =  40 

6X    8  =  48 

7X   8  =  56 

8X   8  =  64 

5X   9.=  45 

6X   9  =  54 

7x    9  =  68 

8X   9=72 

5x10  =  50 

6X10  =  60 

7x10  =  70 

8X10  =  80 

5X11=55 

6X11=66 

7x11  =  77 

8X11=88 

5X12  =  60 
9X  0=     0 

6X12  =  72 

7  X  12  =  84 

8X12  =  96 

10  X  0=     0 

11 X  0=     0 

12X  0=     0 

9X   1=     9 

10  X   1=  10 

11 X    1=   11 

12X   1=   12 

9X  2=   18 

10  X  2=  20 

11  X   2=  22 

12 X   2=  24 

9X  8=  27 

10  X  8=  80 

11  X  8=  88 

12 X   8=  86 

9X  4=  86 

10  X  4=  40 

11  X  4=  44 

12  X  4=  48 

9X  5=  45 

10  X  5=  50 

11  X   5=  55 

12  X  5=  60 

9X  6=  54 

10  X  6=  60 

11  X  6=  66 

12 X  6=  72 

9X   7=  68 

10  X  7=  70 

11  X  7=  77 

12 X  7=  84 

9X  8=  72 

10  X  8=  80 

11  X  8=  88 

12 X   8=  96 

9X  9=  81 

10  X  9=  90 

11  X   9=  99 

12  X  9  =  108 

9x10=  90 

10x10=100 

11X10=110 

12x10=120 

9X11=  99 

10X11  =  110 

11X11  =  121 

12X11  =  182 

9X12  =  108 

10X12  =  120 

11X12=182 

12X12=144 

ADVANCED   ARITHMETIC. 


28 


13.  Process  of  Multiplication. —  Place  the  multi- 
plier under  the  multiplicand ;  multiply  by  each  figure  of  the 
multiplier,  beginning  with  units.  Always  place  the  right-hand 
figure  of  each  product  under  that  figure  of  the  multiplier  used 
in  obtaining  it.     Adxi  these  products  for  the  complete  product. 


EXAMPLES. 


1.  Multiply  379  by  67. 


Process. 
379 
67 
2653 
2274 
25393,  result. 


Explanation:  (1)  7x9  =  63.  Write  the  3  below  and 
carry  the  6.  7x7  =  49.  49  +  6  (carried)  =55.  Write 
the  5  (units)  below,  and  carry  the  5  (tens).  7x3  =  21. 
21  +  5  (carried )  =  26.     Write  the  26  below. 

(2)  Multiply  by  the  6  in  the  same  way. 

(3)  Add  these  products.  The  complete  product  is 
25393. 


Note. — The  6  of  the  multiplier  is  6  tens,  and  the  first  figure  of  its 
product,  which  is  4,  is  also  4  tens.  Therefore  the  4  is  placed  under  the 
6,  or  in  the  second  order. 

Note.— The  ''Long  Way*'  given  in  the  following  examples  is  for  ex- 
planation, and  not  to  be  practiced  by  the  pupil. 


2.  Multiply  57  by  40. 
Long  Way. 

4Q  Note.— In  the  Short  Way,  we  place  the  4 

under  the  7,  and  the  0  to  the  right  of  the  4. 
Bring  down  the  0  and  multiply  by  4  only. 


00 
228 


Short  Way, 

57 
40 


2280 


2280 


3.  Multiply  870  by  500. 

Long  Way. 

Note. — In  the  Short  Way,  we  place  the  5 
under  the  7,  letting  the  O's  fall  to  the  right. 
Multiply  by  the  5,  and  bring  down  three  O's 
—  one  for  the  0  in  the  multiplicand,  and  two 
for  the  O's  in  the  multiplier. 
185000 


Short  Way. 


370 

500 


185000 


24  ADVANCED    ARITHMETIC. 

U.  Multiply  246  by  502. 

Long  Way.  Short  Way. 

246 

gQ2  Note. — We  need  not  multiply  by  the  0  in  ^aq 

— —-  the  multiplier.    But  we  must  be  careful  to  502 

QQQ  put  the  0  of  the  partial  product  obtained  by  7^ 

1230  multiplying  by  the  5  under  the  5.  1230 

123i92  123492 

5.  Multiply  678  by  1200. 

678 
1200  Note.— Multiply  by  12,  not  by  2  and  1. 

813600 

Tests  of  Accuracy:  (1)  Review  the  work,  or  (2)  multiply  the 
multiplier  by  the  multiplicand,  considering  the  multiplicand  abstract. 

EXERCISE  VII. 

Find  products  : 

1.  400x583*  16.   724x6743 

2.  530x475  17.   594X3948 

3.  640X873  18.   475x9873 
i.  370x582  19.   649X8479 

5.  420x897  20.   709x4377 

6.  730X948  21.   806x9835 

7.  890x345  22.   639x3829 

8.  900x760  23.   738x5893 

9.  730x800  2J^.   1263x85643 

10.  1200X548  25.  2095x56789 

11.  1100X675  26.  4075X83064 

12.  246X375  27.  8764x93725 

13.  856x948  28.  4609x85274 
U,  987x987  29.  5007x95683 
15.   876X955  30.  4199x87878 

*The  multiplier  Is  placed  before  and  the  multiplicand  after  the  sign  "  X-" 


ADVANCED   ARITHMETIC.  25 

31.  Multiply  504768  by  943000  J^l.  75x94x237 

32.  Multiply  724256  by  23462  J^2.  184x276x329 
S3.  Multiply  958740  by  37542  Jf.S.  101x279x834 
3J^.  Multiply  818475  by  30724  U-  642x200x730 

35.  Multiply  798648  by  48056  J^5.  820x540x422 

36.  Multiply  829874  by  58876  J^6.  25  X  84  X  68  X  75 

37.  Multiply  941072  by  86754  ^7.  38  X  240  X  820  X  97 

38.  Multiply  480762  by  10000  J^8.  39x800x600x240 

39.  Multiply  708840  by  25000  J!^9.  42  X  560  X  720  X  980 
1^0.  Multiply  984682  by  92845  50.  125x836x457x789 

F.    DIVISION  OF  INTEGERS. 

14.  Definitions,  Signs,  and  Principles. — Divi- 
sion is  the  process  of  finding  how  many  times  one  number 
contains  another.  The  number  contained  is  called  the  Divi- 
sor. The  number  containing  the  divisor  is  called  the  Divi- 
dend.    The  result  of  division  is  called  the  Quotient. 

The  Sign  of  Division,  -^,  is  read  divided  by.  The 
Sign  of  Ratio,  : ,  which  is  also  a  sign  of  division,  is  read 
the  ratio  of. ...to 

Principles:  1.  The  dividend,  divisor  and  remainder  must  be 
similar  numbers. 

2.  The  quotient  must  be  abstract. 

3.  The  dividend  equals  the  'product  of  the  divisor  and  quotient, 
plus  the  remainder. 

Note.— Many  authors  hold  that  the  divisor  may  be  abstract  and  the 
quotient  similar  to  the  dividend.  Such  a  principle  is  inconsistent  with 
the  above  definition  of  division.  The  author  believes  that  the  above 
definition  is  broad  enough  to  cover  every  application  of  division.  (See 
Art.  16.) 

35^5  =  7 

is  read  35  divided  by  5  equals  7. 


26  ADVANCED    ARITHMETIC. 

35:5  =  7 
is  read  the  ratio  of  35  to  5  equals  7.    Each  means  that  35  things  contain  5 
of  those  things  7  times.     ( See  further  explanation,  Art.  17.)    The  35  is  the 
dividend  ;  the  5,  the  divisor  ;  and  the  7,  the  quotient. 

15.  The  Process  of  Short  Division. —  In  Short 
Division,  the  process  is  performed  mentally,  and  the  dividend, 
divisor,  and  quotient  only  are  written.  Short  division  should 
be  employed  when  the  divisor  does  not  exceed  12. 

Put  the  divisor  on  the  left  of  the  dividend,  and  begin  at  the 
left  of  the  dividend  to  divide.     Write  the  quotient  below. 

It  would  be  impossible  to  divide  the  whole  of  a  large  divi- 
dend at  a  glance ;  but  the  number  represented  by  the  first  one 
or  two  figures  on  the  left  can  be  divided ;  the  remainder,  if  any 
from  this  division,  can  be  reduced  to  the  next  lower  order  and 
the  division  continued  until  all  figures  of  the  dividend  are  used. 
The  following  examples  explain : 

EXAMPLES. 

i.  Divide  76  by  4. 

Process.  Explanation:  (1)  7  +  4  =  1,  and  3 remaining.    Write  the  i 

4)76  below  the  7. 

19  (2)  The  remainder  3  is  tens.    3  tens  +  6  units  =  36  units. 

36  +  4  =  9.    Write  the  9  below.     Quotient  19. 

2.  Divide  169  by  8. 

Process.  Explanation :    (1)  1  of  the  dividend  is  smaller  than  8, 

8)169  then  we  use  16. 

21  (2)  16-^-8  =  2.    Write  the  2  below  the  6. 

i-gui  1  (3)  9  +  8  =  1,  rem.  1.    Write  the  quotient  i  to  the  right 

of  the  2,  and  the  rem.  1  below.     Quotient  21,  remainder  1. 

3.  Divide  7220  by  6. 

Process.  Explanation  :  (1)  7-*- 6  =  1,  rem.  1.     AVrite  the  quotient  1 

6)7220  below  the  7. 

1203  (2)  Remainder  1  is  1  thousand.     1  thousand +2  hundred 

rem.  2  =12  hundred.     12  +  6  =  2.    Write  the  2  below. 

(3)  2  +  6  =  0,  rem.  2.     Write  0  below. 
(4)  The  remainder  2  tens  =  20  units.     20  +  6  =  3,  rem.  2.    Write  the  3  at 
the  right  of  the  0,  and  the  rem.  2  below.     Quotient  1203,  remainder  2. 


ADVANCED   ARITHMETIC. 


27 


Tests  of  Accuracy:    (1)  Review  the  work,  or  (2)  apply  prin- 
ciple 3. 

EXERCISE  YIIL 

Copy  and  find  quotients  and  remainders  : 

1.  2.  3.  4.  5, 

8)828  6)372  9)j405  7)595  5)895 


6, 
4)4567 

11. 

10)4892 


7. 
6)8946 

12, 
11)5948 


8. 
7)5938 

13. 
12)8736 


9. 
3)9874 

U. 

11)9878 


10. 
9)8064 

15. 

10)5306 


Find  quotients  and  remainders  : 


16.  84695 -T- 11 

17.  59060-^10 

18.  73421-^12 

19.  12345^12 

20.  23456^9 


21.  73024-T-8 

22.  16295-^ll 

23.  18020^6 
21^.  16200-^12 
25.  90000^7 


26.  235061-^12 

27.  456789 --11 

28.  760890^9 

29.  888888 -^  12 

30.  987654-f-8 


16.  Tliree  Applications  of  Division. 

9)36 

4 

The  fact  of  this  division  is,  that  36  things  contain  9  of  those 

things  4  times. 

Note. — We  must  be  careful  not  to  fall  into  the  error  of  thinking  that 
36  contains  9  as  a  bucket  contains  water.  The  word  ^^contains,''  as  here 
used,  means  ^Hs  made  up  of,"  or  *H*s  composed  of." 

First  Application. — Problem  :  There  are  36  apples  in  a  lot. 
How  many  times  can  we  take  9  apples  from  the  lot  ? 

Answer :  From  division  we  know  that  36  apples  contain  9 
apples  4  times.  Then,  we  can  take  9  apples  from  the  lot  4 
times. 

This  is  the  most  direct  application  of  division,  and  the  only 
one  which  we  have  presented  so  far  in  this  part. 


28  ADVANCED   ARITHMETIC. 

Second  Application. — Problem:  There  are  86  apples  in  one 
lot  and  9  apples  in  another.  The  first  lot  is  how  many  times 
as  large  as  the  second  lot  ? 

'  Answer  :  From  division,  we  know  that  36  apples  contain  9 
apples  4  times.  Then,  36  apples  must  be  4  times  as  large  as  9 
apples  of  the  same  size. 

This  is  the  application  nearest  like  the  direct  application, 
and  may  be  called  the  Hatio  Idea  of  Division. 

Third  Application. — Problem  :  There  are  36  apples  in  a  lot. 
What  is  i  of  the  lot  ? 

Answer:  From  division,  we  know  that  36  apples  contain  9 
apples  4  times.  Suppose  we  take  9  of  the  36  apples,  and  with 
them  begin  9  new  lots ;  for  every  time  we  can  take  9  apples 
from  the  36  apples,  we  can  put  1  apple  in  each  new  lot.  But 
36  apples  contain  9  apples  4  times.  Therefore  we  can  put  4 
apples  in  each  new  lot.  Thus,  we  have  separated  36  apples 
into  9  equal  new  lots,  and  one  lot  contains  ^  of  36  apples,  or 
4  apples. 

Note. — If  necessary,  the  teacher  should  illustrate  this  with  objects. 
This  may  be  called  the  Fraction  Idea  of  Division. 

17.  Different  Ways   of  Expressing  Division. 

There  are  many  ways  or  methods  of  expressing  division.     The 
following  are  in  common  use : 

(1)  86-^9 

This  expression  is  usually  read,  ^^36  divided  by  9."  It 
means  (1)  36  contains  9,  (2)  the  ratio  of  36  to  9,  or  (3)  i  of 
36.    It  is  a  general  method  of  expressing  division. 


ADVANCED   ARITHMETIC. 

(2)  Y 

This  expression  may  be  read,  (1)  ''S6  divided  by  9,"  (2) 
''The  ratio  of  36  to  9,"  or  (3)  ''36  ninths.''  It,  like  the  first 
expression,  is  a  general  method  of  expressing  division. 

■^       (3)  36:9 

Read,  "The  ratio  of  36  to  9."  This  expresses  only  the 
Ratio  Idea  of  Division. 

(4)  i  of  86 

Read,  "One-ninth  of  36."  This  expresses  only  the  Frac- 
tion Idea  of  Division.  ' 

18.  Disposing  of  the  Remainder. 

EXAMFIiE. 

1.  Divide  134  by  5. 

5)134 


26t 

Explanation  :  After  we  had  obtained  the  quotient  26,  we 
had  a  remainder  of  4.  This  4  is  to  be  divided  by  5.  But  we 
learned  in  Expression  (2),  Article  17,  that  4  divided  by  5 
may  be  put  in  the  form,  |,  and  called  "four-fifths."  This 
we  put  on  the  right  of  the  26  and  call  the  expression,  26|, 
"twenty-six  and  four-fifths."    26 1  is  the  exact  quotient. 

When  the  exact  quotient  is  an  integer,  the  dividend  is  said 
to  be  divisible  by  the  divisor.     Thus, 

36--9=4. 
36  is  divisible  by  9. 


EXERCISE 

IX. 

Find  exact  quotients  : 

7.  347-f-7                  6.  834:7 

11.  }  of  965 

16. 

H^ 

2.  545 --9                  7.  209:6 

12.  i  of  677 

17. 

-w 

3.  806--11                 8.  707:11 

13.  yV  oi  509 

18. 

H^ 

4.  793-r-12                9.  651:5 

U.  tV  of  B27 

19. 

iOi 

5.  820-^9                10.  939:  10 

15.  ^ij  of  436 

20. 

-w 

ADVANCED   ARITHMETIC. 


19.  Ijong  Division. —  In  Long  Division,  the  process  is 
the  same  as  in  short  division,  but  the  work  is  written.  Write 
the  quotient  above  or  to  the  right  of  the  dividend.  The  author 
believes  that  placing  the  quotient  above  has  some  advantages. 


EXAMPIiBS. 


1,  Divide  464  by  17o 
Process. 

27 
17)464               Explanation:    (1)  46  +  17  =  2+.*     2x17  =  34. 
34             Bring   down    4.      124  +  17  =  7  +  .      7x17  =  119. 
124           Quotient  27,  remainder  5. 
119 

46-34  =  12. 
124-119  =  5. 

5  rem. 

Since  we  must  now  deal  with  large  divisors,  it  is  not  always 
easy  to  tell  just  what  the  quotient  figure  should  be  without 
trying.  The  following  examples  will  show  how  to  find  the 
quotient  figure  by  trial. 

Ig 

23)387  Explanation  :  Suppose  we  do  not  know  how  many  times 

23  23  is  contained  in   157,  and  we  try  5  times.    5x23  =  115, 

1^  157-115  =  42.    But  42  is  larger  than  23;   therefore,  157 

115  will  contain  23  more  than  5  times. 

Remember  :  When  the  remainder  is  larger  than  the  divisor  the 
quotient  figure  is  too  small. 

— -  Explanation :    If  we  do  not  know  how  many  times  157 

23)387  contains  23,  suppose  we  try  7  times.    7  x 23  =  161.    But  161 

is  larger  than  157;    therefore,  157  will  not  contain  23 


23 

157  7  times. 

161 


Remember  :  When  the  number  to  be  subtracted  is  larger  than  the 
number  from  which  you  are  to  subtract,  the  quotient  figure  is  too 
large. 


*Bead,  2+,  "ficoplus."    It  here  means  2  and  a  remainder. 


ADVANCED   ARITHMETIC.  31 

.    2.  Divide  101655  by  251. 
Process. 

^Q5  Explanation :  After  bringing  down  the  first  5  of  the  div- 

25l)  101655  idend,  the  125  thus  formed  is  too  small  to  contain  the 

IQQ^  divisor,  251.     Put  a  0  in  the  quotient,  bring  down  the 

1255  next  figure,  and  proceed  as  before.     Quotient,  405. 
1255 

S.  Divide  3400  by  200. 

Explanation :  Out  off  all  O's  found  on  the  right  of  the 

^^^^^^'  divisor,  and  an  equal  number  of  figures  from  the  right 

-li —         of  the  dividend.   Divide  the  remaining  part  of  the  div- 

2x00)  34x00         idend  by  the  remaining  part  of  the  divisor.  Quotient,  17. 

4.  Divide  3405  by  200. 
Process. 

17^^  Explanation :    The  5  cut  off  is  remainder.     Exact 

2yS)0)3i^b  quotient,  mh. 

5  rem. 

5.  Divide  3505  by  1700. 
Process. 

2^7^,^  Explanation :  If  there  be  a  remainder  from  the  di- 

17  00)35  05  viding,  the  part  of  the  dividend  cut  off  is  annexed  to 

34  form  the  complete  remainder.    Exact  quotient,  2^.?^js. 

105  rem. 

6.  Divide  84050  by  1000. 

Process.  Explanation  :  When  the  divisor  is  10,  100,  1000,  etc., 

34rt^iy  the  part  cut  off  of  the  dividend  is  remainder,  and  the 

1x000)34x050  P*^*  ^ot  cut  off  is  the  integral  quotient.    Exact  quo- 

50  rem.  tient,  34j^^j5. 

EXERCISE  X. 

Find  quotients  : 

1.  8547-^37  7.  45368--106 

2.  15170-^74  8.   111366^207 

3.  388564-92  9.  176868--306 

4.  48418-^86  10.  258210--342 

5.  373520--580  11.  150181---179 

6.  597760--640  12.  177156-^259 


32  ADVANCED   ARITHMETIC. 

13.   510816-^554  19.   429586-^347 

U.   324264^-458  20.   355074-^249 

15.  481866--539  21.  715464-^456 

16.  363636-^468  22.  800197-f-673 

17.  514201-f-579  23.  11304202^1729 

18.  316386^378  2Jf..  100812054-^5649 

Find  integral  quotients  and  remainders  : 

25.  8581-^231  30.  600181-^937 

26.  15317-^206  31.  45979^433 

27.  39077-^370  32.  112559-^538 

28.  49092-7-563  33.  176868-^578 

29.  374813-r-644  3J^.  324264-^709 

Find  exact  quotients : 

^5.  348657-^1000  J^O.  98764379 -^  137000 

36.  4897531 -- 10000  J^l.  13951475^958 

37.  7306477^45000  1^2.  32657727 -^  75632 

38.  8907601 ---10800  I^3.  7391555011^8604 

39.  325093 -=-72000  U-  26083952680^36745 

20.  General  Review. 

EXERCISE  XI. 

1.  Define  addition,  addend,  sum. 

2.  Give  the  principles  governing  addition. 

3.  Using  the  following  addends,  explain  in  full  the  process 
of  addition : 

24750,  84723,  203079. 

4.  Mr.  Jones  has  real  estate  worth  $7560;  personal  property, 
such  as  household  goods  and  stock,  worth  $1739;  a  stock  of 
merchandise  worth  $8420;  and  a  bank  deposit  of  $3124.  How 
much  is  Mr.  Jones  worth  altogether  ? 


ADVANCED    ARITHMETIC.  ^6 

5.  The  sum  of  7084,  537,  9468,  and  one  other  number  is 
28221.     Find  the  other  number. 

6.  Explain  fully  the  process  of  subtraction,  using  the  follow- 
ing  example:  3408- 1786=  (     )? 

7.  What  is  subtraction  ?     Subtrahend  ?    Minuend  ?     Differ- 
ence ? 

8.  Give  the  principles  governing  subtraction. 

9.  A  man  worth  $8750  lost   by  fire  a   house  worth  $1280. 
What  was  he  then  worth  ? 

10.  The  remainder  is  756,  and  the  subtrahend,  3146.     Find 
the  minuend. 

11.  The  minuend  and  remainder  are  14255  and  6728  respect- 
ively.    Find  the  subtrahend. 

12.  Explain  fully  the  process  of  multiplication,  using  the 
following  example:       72x8407=(     )? 

IS.  Define  multiplication,  multiplicand,  multiplier,  product. 
14..  Give  the  principles  governing  multiplication. 

15.  346  X(     )  =  195144? 

16.  (     )X  730 =88330? 

17.  A  has  246  cattle ;  but  B  has  26  times  as  many  as  A.   How 
many  cattle  has  B  ? 

18.  Two  numbers   multiplied    together  give   a   product   of 
463686 ;  one  of  the  numbers  is  654.     Find  the  other  number. 

19.  Define  division,  divisor,  dividend,  quotient,  remainder, 
short  division,  long  division. 

20.  Give  the  principles  governing  division. 

21.  Explain  fully  the  process  of  short  division,  using  the 
following  example:       753646^9=  (     )? 

22.  Explain  fully  the  process  of  long  division,  using  the  fol- 
lowing  example:  igr,i4^^^^(^     )y 


34  ADVANCED   ARITHMETIC. 

23.  By  what  rule  do  you  know  when  the  quotient  figure  is 
too  large  ? 

24.  By  what  rule  do  you  know  when  the  quotient  figure  is 
too  small  ? 

25.  How  many  $1000-shares  in  a  capital  stock  of  $2000000  ? 

26.  194788-^  (     )  =  281? 

27.  (     )-f- 1248= 821? 

28.  289761  ^  (     )  =  875,  remainder  186  ? 

29.  74475-^824=229,  remainder  (     )? 

30.  How  do  you  shorten  the  process  of  multiplication  when 
there  are  O's  on  the  right  of  the  multiplier  or  multiplicand  ? 

31.  How  do  you  shorten  the  process  of  division  when  there 
are  O's  on  the  right  of  the  divisor  ? 

32.  A  and  B  start  from  the  same  place:  A  goes  east  560 
miles;  B  goes  west  489  miles.  Find  the  distance  between 
them. 

33.  A  and  B  start  from  the  same  place  and  travel  in  the 
same  direction — A,  1846  rods;  B,  859  rods.  Find  the  distance 
between  them. 

34.  What  is  the  number,  from  which,  if  784  be  taken,  the 
remainder  is  591  ? 

35.  What  is  the  number,  to  which,  if  7428  be  added,  the  sum 
will  be  24178? 

36.  Find  the  number,  from  which,  if  480  be  subtracted,  the 
remainder  divided  by  79,  the  quotient  will  be  24. 

37.  What  number  is  that,  from  which,  if  765  be  subtracted, 
the  remainder  multiplied  by  184,  and  1795  be  added  to  the 
product,  the  sum  will  be  288347  ? 

38.  If  875  be  subtracted  from  the  difference  between  two 
numbers,  126  will  remain.  895  is  the  smaller  number;  find 
the  larger. 

39.  Four  men  have  $7500:  the  1st  has  $1500;  the  2d,  $1284; 
the  3d,  $2179.     How  much  has  the  4th  ? 


ADVANCED   ARITHMETIC.  35 

Jfi.  Find  the  sum  of  four  numbers,  if  the  1st  is  177,  the  2d 
is  316  more  than  the  1st,  the  3d  is  741  more  than  the  2d,  and 
the  4th  is  658  less  than  the  3d. 

G.    SIMPLIFYING  NUMERICAL  EXPRESSIONS. 

21.  A  Term. — When  a  numerical  expression  is  separated 
into  parts  by  either  of  the  signs,  plus  (  +  )  or  minus  (  — ),  these 
parts  are  called  Terms  ;  when  not  so  separated,  the  expres- 
sion itself  is  a  term.     Thus, 

5-2+7=10 
5,  2,  7,  and  10  are  terms. 

Sometimes  a  term  is  composed  of  two  or  more  numbers  joined 
together  by  signs,  X  or  -^.     Thus, 

3x4-8-^2 

3x4,  and  8  +  2,  are  terms. 

Two  or  more  terms  may  be  inclosed  by  parentheses,  (  ),  and 
considered  as  a  single  term.     Thus, 

(4+5) -(10-4+2) 
(4+5)  and  (10-4+2)  are  terms. 

When  an  expression  in  parentheses  is  to  be  multiplied  by  a 
number,  that  number  is  usually  placed  just  before  the  paren- 
theses.     Thus,  tr/A   ,  rT\       Atr 

'  5(4+5)  =45 

5(4+5)  means  5  times  the  sum  of  4  and  5y  or  5x9. 

A  dividend  or  a  divisor  may  be  composed  of  two  or  more  num- 
bers joined  by  signs.     Thus, 

^g±|)=5(4+5)-^(12-8). 

These  two  expressions  are  equal  as  indicated.  The  dividend  is  5  times 
the  sum  of  4  and  5,  and  the  divisor  is  the  difference  between  12  and  3. 
Each  expression  is  a  term. 


do  ADVANCED    ARITHMETIC. 

22.  Process. — In  simplifying  a  numerical  expression, 
(1)  always  simplify  each  term  by  itself  and  (2)  add  or  sub- 
tract the  terms  as  indicated  by  the  sign',  plus  or  minus. 

EXAlCFIiES. 

1.  Simplify  7-(9-5)+4x5-20-^4=(     )? 

(1)  7-(9-5)+4x5-20-^4. 

(2)  7-4+20-5  =  18,  result. 

Explanation  :  1st  term,  7;  2d  term,  9-5  or  4;  3d  term,  4x5  or  20; 
4th  term,  20-»-4  or  5. 

2.  Simplify  5(12-10)+4x6-(2+9-8). 

(1)  5(12-10)+4x6-(2+9-3). 

(2)  10+24-8  =  26,  result. 

3.  8(9--4)  _27h-9+3(5+3-7)  =  (     )? 

(1)  8(9-4)  _27h-9+3(5+3-7). 

(2)  4-3+3  =  4,  result. 

EXERCISE  XII. 

Simplify : 

1.  4-8+10-5+12 

«.    12-^4+7-6+14 

3.  24-6x3+4x5x3 

4.  74-10x7+8x3 

5.  8-(4-2)  +  (9-3)-(8-5) 

6.  9+(7+l)+2(10-4)+5x6 

7.  8+7  ,21-32(7+3) 
10-5"^  4+2        9-5 

8.  5x8-^i^±^+(25+5)-4-(ll-5) 

^-  5^+g^^-7+8x5-7(12-8+4-2) 
^^-  |J|^+64-^(8+5)-(5+ll-4) 


ADVANCED   AKITHMETIC. 
H.    DEBIVED  OFEBATIONS. 

1.     FACTORINO. 


87 


23.  Definitions  and  Principles. — A  Prime 
Number  is  an  integer  that  cannot  be  formed  by  multiplying 
two  or  more  other  integers  together.  Below  is  a  table  of  prime 
numbers,  to  be  used  for  reference: 


TABLE  OF  PRIME  NUMBERS  FROM  1  TO  1000. 

1 

59 

189 

288 

887 

489 

557 

658 

769 

888 

2 

61 

149 

289 

847 

448 

568 

659 

778 

887 

8 

67 

151 

241 

849 

449 

569 

661 

787 

907 

5 

71 

157 

251 

858 

457 

571 

678 

797 

911 

7 

78 

168 

257 

859 

461 

577 

677 

809 

919 

11 

79 

167 

268 

867 

468 

587 

688 

811 

929 

13 

88 

178 

269 

878 

467 

598 

691 

821 

987 

17 

89 

179 

271 

879 

479 

599 

701 

828 

941 

19 

97 

181 

277 

888 

487 

601 

709 

827 

947 

28 

101 

191 

281 

889 

491 

607 

719 

829 

958 

29 

108 

198 

288 

897 

499 

618 

727 

889 

967 

81 

107 

197 

298 

401 

508 

617 

788 

858 

971 

87 

109 

199 

807 

409 

509 

619 

789 

857 

977 

41 

118 

211 

811 

419 

521 

681 

748 

859 

988 

48 

127 

228 

818 

421 

528 

641 

751 

868 

991 

47 

181 

227 

817 

481 

541 

648 

757 

877 

997 

58 

187 

229 

881 

488 

547 

647 

761 

881 

A  Composite  Number  is  an  integer  that  can  be  formed 
by  multiplying  two  or  more  other  integers  together.     As, 

4,        6,        8,         9,         10,         12,         15,         16. 

An  Even  Number  ends  in  0,  2,  4,  6,  or  8;  all  others  are 
Odd  Numbers. 

Factors  of  a  number  are  those  numbers  which,  multiplied 
together,  form  that  number.     As, 

8x5x7  =  105. 
3,  5,  and  7  are  factors  of  105. 


38  ADVANCED   ARITHMETIC. 

A  Prime  Factor  is  a  prime  number  that  is  a  factor. 
A  Composite  Factor  is  a  composite  number  that  is  a 
factor.     Thus,  8x4=12 

3  is  a  prime  factor  of  12 ;  4  is  a  composite  factor  of  12. 

A  Common  Factor  of  two  or  more  numbers  is  a  factor 
of  each  of  them.     Thus, 

7X2  =  14;  7X3=21. 
7  is  a  common  factor  of  14  and  21. 

Two  numbers  that  have  no  common  integral  factor  (except 
one)  are  prime  to  each  other.     Thus, 

3x7=21 

5X11=55 

21  and  55  are  prime  to  each  other. 

A  number  can  have  but  one  set  of  prime  factors,  but  a  num- 
ber that  has  three  or  more  prime  factors  (  besides  one)  may 
have  more  than  one  set  of  factors,  some  of  which  are  compos- 
ite.     Thus,  30=3X2X5; 

but  80=3X10=2X15=6X5. 

30  has  but  one  set  of  prime  factors,  2,  3,  5;  but  it  has  three  sets  of 
factors,  some  of  which  are  composite,  3,  10 ;  2,  15 ;  6,  5. 

A  Multii^le  of  a  number  is  a  number  that  contains  that 
number  an  integral  number  of  times.     Thus, 

24-T-12=2. 

24  is  a  multiple  of  12. 

A  Common  Multij^le  of  two  or  more  numbers  is  a 
number  that  is  a  multiple  of  each  of  them.     Thus, 

24^6=4.  24-^8=3. 

24  is  a  common  multiple  of  6  and  8. 


ADVANCED   ARITHMETIC.  39 

A  Divisor  of  a  number  is  a  number  that  will  divide  that 
number  and  give  an  exact  integral  quotient.     Thus, 

4  is  a  divisor  of  24. 

A  Common  Divisor  of  two  or  more  numbers  is  a  num- 
ber that  is  a  divisor  of  each  of  them.     Thus, 

24-f-6=4. 

42-^6^:7. 

6  is  a  common  divisor  of  24  and  42. 

General  Principles:  1.  The  multiplier  and  multiplicand  are 
factors  of  the  product. 

2.  The  divisor  and  the  exact  quotient  are  factors  of  the  dividend. 

3.  Any  composite  number  is  equal  to  the  product  of  all  its  prime 
factors. 

Jf..  All  of  any  set  of  factors  of  a  concrete  number  must  be  abstract^ 
except  one  factor ^  which  must  be  concrete  and  similar  to  the  number 
itself. 

Note. — It  is  immaterial  which  factor  is  considered  concrete.  For  ex- 
ample, factor  the  number  30  bushels. 

2x3x5  bushels  =  30  bushels. 
2  X  5  X  3  bushels  =  30  bushels. 
5x3x2  bushels  =  30  bushels. 
2x5x3x1  bushel  =  30  bushels. 

5.  A  factor  of  a  number  is  a  factor  of  any  multiple  of  that 
number. 

6.  A  factor  of  any  two  numbers  is  also  a  factor  of  their  sum  and 
their  difference. 

7.  When  a  number  is  divided  by  one  of  its  prime  factors^  the  quo- 
tient is  the  product  of  all  the  remaining  prime  factors  of  the  number. 

Principles  to  be  Used  in  Inspecting  a  Number  for  Factors  : 

1.  One  factor  of  a  number  is  2^  if  the  number  ends  in  0, 2^  -^,  6,  or  8. 


40  ADVANCED   ARITHMETIC. 

2.  One  factor  of  a  number  is  3,  if  3  is  a  factor  of  the  sum  obtained 
by  adding  the  digits  of  the  number. 

3.  One  factor  of  a  number  is  4,  if  4-  is  a  factor  of  the  number  ex- 
pressed by  the  two  right-hand  figures. 

4-.  One  factor  of  a  number  is  5,  if  the  number  ends  in  0  or  5. 

5.  One  factor  of  a  number  is  6,  if  2  and  3  are  factors.  (  See  prin- 
ciples 1  and  2.) 

6.  One  factor  of  a  number  is  7,  if  7  is  a  factor  of  the  sum  of  once 
the  first  figure^  plus  3  times  the  second,  plus  2  times  the  third,  plus 
6  times  the  fourth,  plus  4  times  the  fifth,  plus  5  times  the  sixth,  plus 
once  the  seventh,  plus  3  times  the  eighth,  and  so  on,  following  the 
order  1,  3,  2,  6,  4,  5  times  the  successive  figures.     Thus, 

7  is  a  factor  of  84755. 
For,  by  the  principle, 

1x5  +  3x5+2x7+6x4+4x3  =  70, 
and  7  is  a  factor  of  70. 

7.  One  factor  of  a  number  is  8,  if  8  is  a  factor  of  the  number  ex- 
pressed by  the  three  right-hand  figures. 

8.  One  factor  of  a  number  is  9,  if  9  is  a  factor  of  the  sum  ob- 
tained by  adding  its  digits. 

9.  One  factor  of  a  number  is  10,  if  the  number  ends  in  0. 

Note. — The  1st,  2d  and  4th  of  these  principles  are  most  practical.  The 
3d,  8th  and  9th  are  often  used.  The  5th  and  7th  are  sometimes  used. 
But  the  6th  is  not  used  in  practice ;  for  it  takes  longer  to  apply  the  prin- 
ciple than  to  divide  by  7. 

24.  Factoring  by  Inspection. — When  numbers  are 
small,  they  are  usually  factored  by  inspection. 

EXAMFIiES. 

1.  Find  the  factors  of  9. 

Think:   9  =  3x3.     Call:   "  Factors  3,  3." 

2.  Factor  l60. 


ADVANCED   ARITHMETIC.  41 

EXERCISE  XIII. 

Factor  by  inspection  : 

1.  25  ^.  72  7.   112  10.  210 

2.  40  5.  96  8.  128  ii.  175 

3.  75  (5.\91  5>.  150  12.  240 

25.  Factoring  by  Division, — When  the  number  to 
be  factored  is  large,  division  is  employed  in  finding  the  factors. 
(See  General  Principles  2  and  7.)  The  successive  divisors  are 
(1)  found  by  inspection  (see  Principles  to  be  used  in  Inspect- 
ing) or  (2)  by  trial. 

EXAMPI.ES. 

1.  Find  the  prime  factors  of  210. 

Explanation :   (1)  By  Prin.  4,  5  is  a  factor  of 
5)210  210. 

3)  42  (2)  By  Prin.  2,  3  is  a  factor  of  42. 

2)  14  (3)  By  Prin,  1,  2  is  a  factor  of  14. 

^  .-.  5,  3,  2,  and  7  are  the  factors  of  210. 

2.  Find  the  prime  factors  of  5040.  Find  5  composite  factors. 

Process. 
(1)    2)5040 


2)2520 

2)1260 

2)  630 

5)  315 

3)  63 

3)  21 

7 

1,  2,  2,  2,  2, 

5, 

3, 

3, 

7. 

(2)  2x2  =  4;  2x2x2  =  8;  2x3  =  6; 
2x2x3  =  12;  2x7  =  14. 

Note. — Two  or  more  prime  factors  of  a  number  multipHed  together 
always  give  a  composite  factor  of  that  number. 


42  P  ADVANCED   ARITHMETIC. 

EXERCISE  XIV. 

Find  the  prime  factors  of — 

i.  2304         J!^.  17160  7.  73920  10.  282960 

2.  2835          5.  20304  8.  39375  11.  146146 

.^.  3000          6.  32340  P.  36288  .    12.  1067220 

i^i^ic?  fourteen  factors  for  each  of  the  following  : 

13.  210  15.  1155  17.  770 

i^.  330     .  16.  462  i^.  510 

26.  Factoring  Expressions  Composed  of  Two 
or  More  Terms. — It  is  sometimes  necessary  or  conven- 
ient to  factor  an  expression  consisting  of  two  or  more  terms. 
This  may  be  done  by  dividing  by  a  factor  common  to  all  the 
terms. 

EXAMPIiES. 

1.  Take  the  factor  5  out  of  20+25. 

Process:  20+25  =  5(4+5). 

Explanation :  5  is  contained  in  20  4  times,  and  in  25  5 
times.  Thus,  5  is  one  factor  and  4+5  is  the  other.  5(4+5) 
may  not  only  be  interpreted  as  5  times  the  sum  of  4  and  5,  but 
also  5  times  4  added  to  5  times  5. 

2.  Take  the  factor  5  out  of  25-20. 

Process:   25-20  =  5(5-4). 

Explanation  :  5(5-4)  may  be  interpreted  5  times  the  differ- 
ence between  5  and  4,  or  5  times  5  minus  5  times  4' 

3.  Take  the  common  factor  out  of  49  —  14+28. 

Process:  49-14+28  =  7(7-2+4). 


ADVANCED    ARITHMETIC.  /  43 

EXERCISE  XV. 

Take  out  the  common  factors  : 

1.  24-4  6.  5+20-15 

2.  72+18  7.  18-9+36 

3.  36-16   •.  8.  48-16+64 

4.  57+38  9.  25+30-45 

5.  75-50  10.  14+28-21+7 

27.  Principles  of  Multii)lication  and  Divi- 
sion Relating  to  Factors. —  Factors  are  of  common 
occurrence  in  the  processes  of  multiplication  and  division, 
and  the  following  principles  will  be  of  service: 

Principles:  1.  Multiplying  any  factor  of  a  product  by  a  num- 
ber, multiplies  the  product  by  that  number. 

Example:   Indicate  the  multiplication  of  3x5x4  by  2.     (3 

forms.) 

Forms:    (1)  (2x3)x5x4  =  6x5x4;  or, 

(2)  3x(2x5)x4  =  3xl0x4;  or, 

(3)  3x5x(2x6)  =  3x5xl2. 

2.  Dividing  any  factor  of  a  product  by  a  number,  divides  the 
product  by  that  number. 

Example  :   Indicate  the  division  of  4x6x8  by  2.     (3  forms.) 

Forms:    (1)  (4  +  2)x6x8  =  2x6x8;  or, 

(2)  4x(6  +  2)x8  =  4x3x8;  or, 

(3)  4x6x(8  +  2)  =  4x6x4. 

3.  A  quotient  is  multiplied  by  a  number  (1)  by  multiplying  its 
dividend  by  that  number,  or  (2)  by  dividing  its  divisor  by  that 
number. 

Example  :  Indicate  the  multiplication  of  24^6  by  3.    (2  forms.) 

Forms:    (1)  (3x24)-h6  =  72  +  6  ;  or, 
(2)  24h-(6h-3)  =  24h-2. 

Note.— The  dividend  or  divisor  may  be  expressed  as  a  product  of  two 
or  more  factors. 


44  ADVANCED   ARITHMETIC. 

Example  :  Indicate  the  multiplication  of  ^^  by  3.     (4  forms.) 


Forms  : 

.-,.  (3X12) X5  _  36X5  , 
^^'    9X6     9X6  '  °^' 

,^^    12X(3X5)  _  12X15  .  ^^ 
(2)    9X6      9X6  »  ^^= 

/o^   12X5   _  12X5  . 
^^'  (9-^3)X6    3X6  '  "  ' 

,  .V   12X5   _  12X5 
^  '  9X(6^3)    9X2* 

4-'  A  quotient  is  divided  by  a  number  (!)  by  dividing  its  dividend 
by  that  number,  or  (2)  by  multiplying  its  divisor  by  that  number. 

Example  1 :   Indicate  the  division  of  ^^-  by  8.     (2  forms.) 


Forms:     (l)^  =  |;or, 

^^^  3X2        6* 


12X18 


Example  2  :  Indicate  the  division  of  -^^  by  3.     (4  forms.) 


4X1 
T^  ,-,v   (12^3)X18      4X18 

Forms:     (D '-ixiT- =  ixn  »  °^' 

.„.  12X08^  _  12X6.  ^^ 

^"^^  ~4xii        ixn'  °^' 

,„.      12X18     ^12X18. 
^^^^  (3X4)X11       12X11 '         ' 
/4\      12X18     ^  12X18 
^*'  4X(3X11)        4X33  • 


EXERCISE  XVI. 

Indicate  as  above  the  multiplication  of — 
1.  4X7  by  8.      (2  forms.)  ^.   _^  by  7.      (3  forms.) 

^.  9 X 12  X 3  by  5.     (3  forms.)       S,  ^  by  5.     (3  forms.) 

3.  -V\0-  by  25.     (2  forms.)  6.  ^^  by  3.     (5  forms.) 

Indicate  as  above  the  division  of — 
7.   12X80  by  6.     (2  forms.)  jq.    5^0  ^7  25.     (3  forms.) 

<?.   15X20X45  by  5.    (3forms.)     11,    ^^^  by  12.     (3  forms.) 
9.  ^^  by  12.     (2  forms.)  i^-    ^  by  7.     (4  forms.) 


ADVANCED   ARITHMETIC.  45 

9.     OANCEIiliATION. 

28.  Definition  and   Principle. — Cancellation 

is  a  process  of  shortening  the  work  of  division  by  omitting  the 
same  factors  from  both  dividend  and  divisor. 

Principle  :  Dividing  both  dividend  and  divisor  by  the  same 
number  does  not  affect  the  quotient. 

Note. — Dividing  the  dividend,  divides  the  quotient  (Art.  27,  Prin.  4). 
Dividing  the  divisor,  multiplies  the  quotient  (Art.  27,  Prin.  3).  But, 
when  the  quotient  is  both  multiplied  and  divided  by  the  same  num- 
ber, its  value  is  unchanged. 

29.  Process. — It  is  usual  to  write  the  dividend  or  the 
factors  of  the  dividend  above  a  horizontal  line,  the  divisor  or 
factors  of  the  divisor  below.  Proceed  to  cancel  and  reject  fac- 
tors common  to  both  dividend  and  divisor.  When  all  common 
factors  are  rejected,  divide  the  product  of  the  remaining  factors 
of  the  dividend  by  the  product  of  the  remaining  factors  of  the 
divisor. 

EXAMPIiES. 

1.  Divide  50x96  by  15x8. 

4 
10  n 

Process:  ^12^^40. 

Explanation:  (1)  Cancel  5  out  of  50  and  15,  leaving  10 
above  and  3  below.  (2)  96  contains  8  12  times ;  cancel  8  and 
96,  placing  12  above.  (3)  12  contains  8  4  times ;  cancel  3  and 
12,  and  place  4  above.    (4)  4  x  10  =  40. 

2.  How  many  times  will  48x27x15  contain  9x6x10? 

4 
^33 

Process  :^1^E2^  =  ^^. 

^x^x;p 

Explanation :  (1)  6  in  48  8  times.  Cancel  6  and  48,  and 
place  8  above.    (2)   9  in  27  3  times.    Cancel,  and  write  3 


46  ADVANCED   ARITHMETIC. 

above.     (3)    5  is  common  to  15  and  10.    Cancel,  and  write 

3  above  and  2  below.    (4)  2  in  8  4  times.    Cancel,  and  write 

4  above.    (5)  4x3x3  =  36. 

EXERCISE  XVII. 

i.  17X13X50  5.  625X720X81 

65x10     ~^  ^'  750x729      ~^  ^* 

2.  72X105X160  6.  400X375X338 

24x42x120  ~^   ^'  507X625X80"^   ^* 

S,  200x57x35  7.  2816x16200x8400 

75x70x19  ~^  ^'  528x1512x1280    ~^  ^^ 

4.  441x210x216  5.  5409x1728x840 

126X105X84  ~^  ^*  8640X10217      ~^  ' 


3.     GREATEST  OOMMON  DIVISOR. 

30.  Definition  and  Principles. — The  Greatest 
Copcimon  Divisor  of  two  or  more  numbers  is  the  largest 
number  that  is  contained  in  each  of  them  an  integral  number 
of  times. 

Principles:  1.  Every  factor  of  a  number  is  a  divisor  of  that 
number. 

2.  The  product  of  two  or  more  prime  factors  of  a  number  is  a  di- 
visor of  that  number. 

3.  A  divisor  of  a  number  is  a  divisor  of  any  multiple  of  that 
number. 

4-.  A  common  divisor  of  two  numbers  is  a  divisor  of  their  sum 
and  of  their  difference. 

5.  The  G.  C.  D.  of  two  or  more  numbers  is  the  product  of  all  the 
prime  factors  common  to  the  numbers. 

6.  The  G.  C.  D.  of  two  numbers  is  the  G.C.  D.  of  either  of  them 
and  their  sum,  or  difference. 


ADVANCED   ARITHMETIC.  47 

31.  Process. — When  the  numbers  are  small,  the  G.  C.  D. 
is  usually  found  by  factoring. 

EXAMPLES. 

1.  Find  the  G.  C.  D;  of  24,  48,  and  72. 

Process. 
24  =  2x2x2x3. 
48  =  2x2x2x2x3. 
72  =  2x2x2x3x3. 
2,  2,  2,  3  are  common  factors. 
.'.  2x2x2x3  =  24,  G.O.D. 

The  following  process  is  also  used  when  the  numbers  are 
small : 

2.  Find  the  G.  C.  D.  of  15,  45,  and  60 

Process. 
5)15,45,60 
3)  3,    9,  12 
1,    3,    4 
3  and  5  are  all  the  common  factors, 
.-.3x5  =  15,  G.O.D. 

Note. — In  this  process,  use  only  such  divisors  as  will  divide  all  the 
numbers. 

When  the  numbers  are  so  large  as  not  to  be  easily  factored 
by  inspection,  the  following  process  is  used: 

3.  Find  the  G.  C.  D.  of  254  and  381. 

Process. 

Explanation  :   Since  127  is  the  G.  0.  D.  of 

254)381  127  and  254,  it  is  also  the  G.  0.  D.  of  254  and 

-^^i- — =      381  ( Prin.  6).     .'.  the  required  G  C.  D.  is  127. 
127)254 
254 

Note.— This  plan  or  process  of  finding  the  G.  0.  D.  is  to  divide  the 
larger  number  by  the  smaller ;  then  divide  the  divisor  by  the  remain- 
der ;  then  continue  to  divide  the  last  divisor  by  the  last  remainder,  until 
there  is  no  remainder.    The  last  divisor  is  the  required  G.  0.  D. 


48  ADVANCED   ARITHMETIC. 

^.  Find  the  G.  C.  D.  of  221,  864,  and  5512. 

Process.  Explanation:  (l)TheG.O.D. 

(1)  (2)               of  364  and  5512  is  52.    But  52  is 

15  4           not  a  factor  of  221.     (2)  The  G. 

364)5512  52)221            ^'  ^'  ^^  ^^  and  221  is  13.    Now, 

364  208    4      ^2  contains  all  factors  common 

"7872  ]^q)52      *^  ^^^  ^"^  ^^^2 »  ■'^^  ^^  *^®  °^^y 

1820      7      '  52      factor  common  to  52  and  221. 

Therefore  13  is  the  G.  0.  D.  of 
221,  364,  and  5512. 
.'.  the  required  G.C.  D.  is  13. 


52)364 


EXERCISE  XVIII. 

Find  the  G.  CD.  of— 

1.  86,  48,  72.  6,  1980,  2511. 

2,  96,  120,  1728.  7.  4840,  12504. 

8.  210,  840,  126.  8.  8070,  2149,  614. 

J^.   1022,  2518.  9,   7384,  (M95,  4821. 

5,   1578,  6881.  10.   1860,  1682,  2040,  4080. 

4.     liEAST  COMMON  MUIiTIPIiE. 

32.  Definition  and  Principles. — The  Least 
Common  Multij)le  of  two  or  more  numbers  is  the  small- 
est number  that  contains  each  of  them  an  integral  number  of 
times. 

Principles:  1.  A  multiple  of  a  number  is  divisible  by  that  num- 
ber; and  a  common  multiple  of  two  or  more  numbers  is  divisible  by 
each  of  them. 

2.  A  multiple  of  a  number  contains  all  the  prime  factors  of  that 
number ;  and  a  common  multiple  of  two  or  more  numbers  contains 
all  the  prime  factors  of  each  of  them. 

3,  The  L.  C.  M.  of  two  or  more  numbers  contains  every  prime 
factor  of  the  several  numbers  the  greatest  number  of  times  it  is  found 
in  any  one  of  them. 


ADVANCED    ARITHMETIC.  49 

4..  The  L.  C.  M.  of  two  or  more  numbers  is  a  multiple  of  all  the 
factors  of  those  numbers. 

5.  The  L.  C.  M.  is  the  product  of  one  of  the  numbers  multiplied  by 
the  quotient  obtained  by  dividing  the  other  numbers  by  their  G.  C.  D. 

Note. — The  L.  C.  M.  of  more  than  two  numbers  may  be  found  by  find- 
ing the  L.  C.  M.  of  two  of  them ;  then,  of  that  result  and  the  third  ;  then, 
of  that  result  and  the  fourth,  and  so  on. 

33.  Process. — The  L.  C.  M.  of  two  or  more  numbers  may- 
be easily  found  by  factoring,  when  the  numbers  are  small. 

EXAMPIiES. 

1.  Find  the  L.  C.  M.  of  15,  12,  and  10. 

Process. 
15  =  3x5. 
12  =  3x2x2. 
10  =  5x2. 

The  L.  0.  M.  must  contain  3  once,  5  once,  and  2  twice.    (Prin.  3.) 
.-.  3x5x2x2  =  60,  L. CM. 

Division  is  often  employed  in  finding  the  L.  C.  M. 

2.  Find  the  L.  C.  M.  of  10,  12,  24,  and  30. 

Process. 
2)10,12,24,30 


2) 

5, 

6, 

12, 

15 

5) 

5, 

3, 

6, 

15 

3) 

1, 

3, 

6, 

3 

1,    1,    2,    1 


Note. — In  this  process,  divide  by  any  prime  number  that  will  divide 
two  or  more  of  the  numbers.  Numbers  not  containing  the  divisor  an  in- 
tegral number  of  times  must  be  brought  down  ;  as  5  and  15  in  the  second 
division  above.  When  the  final  results  are  prime  to  each  other,  the  con- 
tinued product  of  all  the  divisors  and  final  results  ( I's  may  be  omitted ) 
will  be  the  L.  0.  M.     ( Prin.  3.) 


50  ADVANCED    ARITHMETIC. 

When  some  of  the  numbers  are  factors  of  others,  these  fac- 
tors may  be  omitted  in  finding  the  L.  C.  M.     (Prin.  4.) 

S.  Find  the  L.  C.  M.  of  21,  15,  and  45. 

Process. 
3)21,  X^,  45  Explanation  :  The   15  may   be 

Z         jT  canceled  from  the  process,  for  it 

3  X  7  X 15  =  315,  L.  C.  M.      '^  ^  ^^^^^  °^  ^^• 

The  L.  C.  M.  of  two  numbers  may  be  found  by  principle  5, 
as  follows : 

^.  Find  the  L.  C.  M.  of  160,  240,  and  280. 

I.  Find  the  L.  0.  M.  of  160  and  240.      II.  Find  the  L.  G.  M.  of  480  and  230. 
(1)  1  (4)  2 


160)240 

160       2 

230)480 

460     11 

80)160 
160 

20)230 
20 

.-.  80  is  G.  0.  D.  of  160  and  240. 

30 

(2)  160-80  =  2. 

20    2 

(3)  2x240  =  480,  the  L.  0.  M.  of 

10)20 

160  and  240. 

20 

.-.  10  is  G.  0.  D.  of  480  and  230. 

(5)230-10  =  23. 

(6)23x480  =  11040,  the    required 

L.C.M. 

EXERCISE  XIX. 

Find  the  L.  C.  M.  of— 

1.  42,  50,  72.  7.  75,  125,  50,  100. 

2.  120,  96,  48.^  8.  105,  84,  63,  147. 

3.  144,  240,  600.  9.  174,  485,  4611,  970. 

4.  256,  120,  860.  10.  264,  144,  824,  576. 

5.  81,  540,  860.  11.  1260,  198,  480,  880. 

6.  24,  36,  48,  60.  12.  2862,  3498,  4158. 


•   ADVANCED   ARITHMETIC.  51 

I.     COMMON  FRACTIONS. 

34.  Definitions  and  Principles. — A  Fraction  is 

a  number  composed  of  fractional  units. 

What  is  a  unit  ?    Integral  unit?    Fractional  unit ? 

A  Common  Fraction  is  a  fraction  expressed  by  two 
numbers,  placed  one  above  and  the  other  below  a  horizontal 
line.     Thus, 

A.  ?1? 

17  9 

The  number  below  the  line  is  called  the  Denominator, 
and  tells  the  size  of  the  fractional  unit.  The  number  above 
the  line  is  called  the  Numerator,  and  tells  the  number  of 
units  in  the  fraction.  The  numerator  and  denominator  are 
called  the  Terms  of  the  fraction. 

Explanation:    The   ''17"  tells  that 
_5_.... Numerator.      the  units    are    seventeenths;    the   "5"  * 

17.. Denominator,      tells  that  this  fraction  expresses  five 
units  — five-seventeenths. 

A  common  fraction  is  also  an  expression  of  division.  Review 
Article  17. 

The  numerator  of  the  fraction  is  the  dividend,  and  the  denom- 
inator, the  divisor. 

When  the  numerator  is  equal  to  or  larger  than  the  denomi- 
nator, the  fraction  is  called  an  Improper  Fraction. 

Principles:  1.  A  fraction  is  multiplied  (1)  by  multiplying  its 
numerator,  or  (2)  by  dividing  its  denominator.  (  See  Prin.  8, 
Art.  27.) 

rLIiUSTRATION. 

Multiply  i  by  3. 

Process:   (1)  3x|  =  ^^^  =  |,  or, 

The  results  agree  ;  for  f  of  a  number  =  J  of  it. 


52  ADVANCED    ARITHMETIC.  » 

2.  A  fraction  is  divided  (!)  hy  dividing  its  numerator ,  or  (2)  by 
multiplying  its  denominator.     (  See  Prin.  4,  Art.  27.) 

HiliUSTRATION. 

Divide  |  by  2. 

Process:    (1)  f  ^2=^  =  |-;  or, 

(2)  Y^^"2X3^  6"- 
These  results  agree ;  for  ^  of  a  number  =  f  of  it. 

3.  The  value  of  a  fraction  is  not  changed  (1)  if  both  its  terms 
are  multiplied  by  the  same  number^  or  (2)  if  both  its  terms  are  di- 
vided by  the  same  number. 

rLIiUSTRATIONS. 

1.  Multiply  both  terms  of  i  by  3. 

r>  3X1        3 

Process:    ^,  =  j- 

But  -g  =  ^.    Therefore,  the  value  is  not  changed. 

2.  Divide  both  terms  of  |  by  2. 

Process  :     g^  =  y 

But  i  =  ^.    Therefore,  the  value  is  not  changed. 

o         o 

35.    Beclucing'  Fractions  to  Hig-lier  Terms.— 

A  fraction  is  reduced  to  higher  terms  by  multiplying  both  its 
terms  by  the  same  number,  according  to  principle  8.  This 
does  not  change  its  value. 

EXAMPLES. 

1.  Reduce  |  to  9ths. 

Process:     y  =  ^  =  9,  result. 

Why  multiply  by  3  ?  Because  the  denominator  3  is  to  be  changed  to 
9.    5x3  =  9. 

Note.— An  integer  may  be  treated  as  a  fraction,  having  1  for  a  denom- 
inator.    Y  =5  or  5=  y.     Y  is  read  5  ones. 


ADVANCED    ARITHMETIC.  53 

2.  Reduce  7  to  Sths. 


Process  :    ^  =  ^-^  =  — ,  result. 

1        5X1        o 


3.  Reduce  |  to  63ds. 


Process -i     (1)  63-5-7  =  9. 


Why  divide  63  by  7  ? 

4.  Reduce  yV  ^^  lOSths. 
Process :  15 

(1)  18)195 


5^  _  9X5  _  45  J 

^"^^    7  -  9X7  -  63'  result. 


\^/    -^'-'/  -^»'"-' 

13 

65 
65 

(2) 

^  =  ^3  =  1^.--"-               . 

Reduce  — 

EXERCISE  XX, 

1.  TAjto48ths. 

6.  W  to  324ths. 

2.  -/y  to  63ds. 

7.  if  to  333d8. 

S.  11  to  37ths. 

5.  9 A  to  2109th8. 

-^.  3t\  to  209thB. 

P.  iJ  to  TOSths. 

5.  if  to  390th8. 

10.  18yVr  to  8181sts 

36.  Reducing  Fractions   to  Lowest  Terms. — 

A  fraction  is  in  its  lowest  terms,  when  its  numerator  and  denom- 
inator are  prime  to  each  other.  Divide  both  numerator  and 
denominator  by  their  G.  C.  D.,  or  reject  all  factors  common  to 
both  numerator  and  denominator.  This  does  not  change  its 
value. 

EXAMPIiES. 

1.  Reduce  f  to  thirds. 

Process  :     j  =  —^  =  -^ ,  result. 

Why  divide  by  2  ?    How  many  3's  in  6  ? 

2.  Reduce  |f  to  lowest  terms. 

Divide  by  G.  C.  D.  of  both  terms.    The  G.  0.  D.  of  12  and 


18  is  6. 


Process  :     |  =  |i|  =  §,  result;  or  ^3  =  ^=  |,  result. 


64  ADVANCED   ARITHMETIC. 

3.  Reduce  y*^^  to  lowest  terms. 
G. CD.  of  49  and  161  =  7. 

n_  49        49-4-7        7  ,,  49       7)  49        7  ,^ 

Process  :    jei  =  i6i:r7  =  23'  ^^^^l*  5  o''  I6I  =  mei  =  23'  ^^^ult. 

EXERCISE  XXI. 

Reduce  to  loivest  terms  — 

^.  A\         ^'  m         7.  m         10.  iff 
^.  Ml         5.  *«         5.  nu        11-  \m 

37.  Reducing  Mixed  Numbers  to  Improper 
Tractions. — Reduce  the  integral  part  to  the  denomination 
of  the  fractional  part  and  add.     The  value  is  not  changed 

EXAMPLES. 

1.  Reduce  8^  to  halves. 

Process  :    (D  3  =  |g  =  |.         (2)  |  +  |  =  |,  result. 
(1)  Reduce  the  integer  to  halves,  (2)  add  the  |. 
Shorter  Form  :    3J  =  ^  +  j  =  |,  result. 

2.  7|  to  4ths. 

Process:    7f  = -j- +  j  =  j,  result. 

3.  Reduce  17-^^  to  an  improper  fraction. 
Mechanical  Work  : 

17 

51  Explanation :    17  reduced  to  twenty-thirds 

34  will  make  23x17,  or  391  twenty-thirds.    Add- 

391  ing  the  9  twenty-thirds,  we  have  400  twenty- 

9  thirds,  or  \Q^. 
400 

Result,  Vs-. 

EXERCISE  XXII. 

Reduce  to  improper  fractions  : 

1.  17-i\  4.  18^  7.  OSiJ.  10.  871A 

2.  24t  5.  75t\  8.  804^  11.  488yViV 

3.  18^^  6.  8941  ^-  132U  i^.  77^AV 


ADVANCED    ARITHMETIC.  55 

38.    Reducing'    Improper   Fractions    to    Inte- 
gers  or  Mixed  Numbers. — This  is  done  by  division. 

See  Article  17.     The  value  is  not  changed. 

EXAMPLES. 

1.  Reduce  ^-^  to  an  integer. 

Process  : 

4)20  Note. — As  an  expression  of  division,  **^0" 

5  result.        is  the  dividend  and  "^  "  the  divisor. 

2.  Reduce  ^W  ^^  ^  mixed  number. 

Process :  1911,  result. 

18)355 

18 


175 
162 


13 
Note. — If  the  fractional  part  of  the  result  is  not  in  its  low- 
est terms,  reduce  by  Article  36. 

EXERCISE  XXIII. 

Reduce  to  an  integer  or  mixed  number  : 

2.  W  5.  -W^-  8.  h^i^-  11,  ^U 

S,  ^-i^  6.  -W/  9.  -\W  12.  1^^^ 

39.  Reducing  Fractions  to  Least  Common 
Denominator. — The  L.  C.  D.  of  two  or  more  fractions  is 
the  L.  C.  M.  of  their  denominators.  The  process  of  reduction 
is  that  of  reducing  fractions  to  higher  terms.  See  Article  35. 
The  values  are  not  changed. 

EXAMPLES. 

1.  Reduce  i,  |,  |- to  L.C.D. 

The  L.  0.  D.  is  12.    Then  the  object  is  to  reduce  the  several 
fractions  to  12ths. 

''   UNIVERSITY 

OF  . 


56  ADVANCED   ARITHMETIC. 

Process 


a)| 

6X1  6 
6X2   12 

(2)| 

_  3X3  _  9 
3X4   12 

<3)| 

_  2X5  _  10 
2X6   12 

Reduce  ^,  -fy,  ^  to  L.  C.  D. 

Process:    A,  A,  U-  (S)  ^^IJl;  7x6=^42, 

6 

^;  5x4  =  20. 

(1)  The  L.  C.  D.  is  72.  (4)  18)72 

72 

~;  11x3  =  33. 

(2)  fi,  fi,  ?|,  result,  ^^^  ^^ 

Note. — The  part  on  the  right,  which  is  the  process  of  find- 
ing the  numerators,  may  be  put  in  such  form  as  is  conven- 
ient. As  soon  as  the  L.  C.  D.  is  found,  the  denominators 
may  be  placed  in  (2)  thus :  f^,  7^,  y^.  Then  find  the  numer- 
ators and  place  them  above. 


EXEKCISE  XXIV. 

Reduce  to  L.  C,  D,  : 

1'  h  A,  ii-  6.  If,  5,  A,  -if. 

2.  ^,  ^,  ^f .  Note.— Reduce  If  to  an  improper 

^,    2.2    X7     11^  fraction  before  commencing  to  find 

/      6     37     13    19  L. CD.    Consider  5  as  f. 

4'  TTj  ftj  -2^?  T¥-  ^     io  1;     HK  ^      1  « 

^'    tt J  tT»  -8  STJ  TT2- 
^-    'sVj  "2^5   Aj    2V- 

40.  Addition  of  Fractions. — Reduce  all  fractions 
to  L.  C.  D.,  add  the  numerators,  and  place  the  sum  over  the 
L.  C.  D. 


ADVANCED    ARITHMETIC.  57 

EXAMPIiES. 

1,  Add  t,  I,  i. 

Process;  l+i  +  \=l%+U  +  io  =  U  =  H^,  result. 

Explanation  :  Mechanical  work  necessary  to  reduce  the 
fractions  to  L.  G.  D.  need  not  be  preserved.  20  thirtieths + 
24  thirtieths +  5  thirtieths  =  49  thirtieths,  or  ff. 

Note. — If  the  result  is  an  improper  fraction,  reduce  it  to  an  integer 
or  mixed  number. 

2.  Add  f ,  I,  H. 

Process:  f  +  f +f|  =  li+IHtl  =  ft  =  2A  =  2f,  result. 

Note. — The  i-^  should  be  reduced  to  its  lowest  terms,  f .    In  results, 
all  fractions  should  be  in  their  simplest  forms. 


3.  Add  64,  lOf ,  5t\. 

Process:    (1)  J+|+t^  = 
fH  1^  =  11 

(2)6t 
101 
5t^ 

221 
Explanation:  The  result  of  adding  the  fractions  is  1|. 
Write  the  |  below  the  fractions,  carry  the  1  and  add  it  to 
the  integers ;  result,  22f . 

EXERCISE  XXV. 
Find  the  value  of— 

2.  181+27^  7.  18A+19tV+22A 

3.  46A+A  8,  8465+17A+T¥ir 

^.  ii+i^+¥k  10.  7if+i7|i+9MJ+24^^f 

41.  Subtraction  of  Fractions. —  Reduce  the  frac- 
tions to  L.  C.  D.  before  subtracting.  Subtract  the  numerator 
of  the  subtrahend  from  the  numerator  of  the  minuend  and 
place  the  remainder  over  the  L.  C.  D. 


58  ADVANCED    ARITHMETIC. 

EXAMPIiES. 

1.  From  f  take  3%. 

Process  :    f  -  A  =  f  e  -  rl  =  ^e »  result. 

Explanation:  20  thirty-sixths  - 15  thirty-sixths  =  5  thirty- 
sixths,  or  ^g.  In  the  result,  always  reduce  the  fraction  to  its 
simplest  form. 

2.  From  17i  take  14^. 

(2)  17^ 
Process  :    (1)  i  -  i  =  A  -  A  =  rV-  14^ 

Si^g,  result. 

3.  From  17|  take  5|. 

(2)  m 

Process:    (1)  f-|  =  ||-f|  =  H-  5| 

11^1,  result. 

Explanation :  Since  |  is  larger  than  f ,  1  must  be  taken  from 
17  and  added  to  the  f ,  making  If  or  ? ;  f  -  f  =  i|.  This  leaves 
only  16  in  the  minuend,  from  which  to  subtract  the  5. 

EXERCISE  XXVI. 

Find  the  value  of — 

1.  if-f  6.  44i-36A 

2.  7i-ii  7.  86||-f| 

^.  18A-«  ^.  538H-26H 

4.  75-fl  P.  452Jo-247Hi 

5.  87^-15^  ^^.  140f|-14-|f 

42.  Multiplication  of  Fractions.— When  the  mul- 
tiplier is  a  fraction,  it  may  be  followed  by  the  word  *'o/"  or 
the  sign,  *'  X."     Thus, 

f  of  10=fXlO. 

Reduce  mixed  numbers  to  improper  fractions  before  multiply- 
ing. 

EXAMPIiES. 

1.  Find  f  of  85. 

Process  Complete :    (1)    ^^  of  35  =  ^  =  5.    (  See  Art.  16,  3d  Ap.) 
3  X  (1)  =  (2)  *  f  of  35  =  3  X  5  =  15,  result. 
Teacher,  show  the  pupil  that  (2)  comes  from  (1)  by  multiplying  by  3. 

*  "3X(1)=(2) "  Is  read  3  times  eqttation  one  equals  equation  two. 


ADVANCED    ARITHMETIC.  59 

5 


Process  Shortened:  ^  =  15,  result. 


2.  Find  |  of  |. 

p           ,                    •  Explanation:     §  is  2  +  3.     Dividing 

1     f  2  _  8     7-  3      «o  ,if  this  by  7  is  the  same  as  dividing  2  by 

r  of  .-^-7-^,  result.  3^7^  ^^21.     2  +  21  =  ^. 

3.  Find  f  of  |. 

Process  Complete:  (1)  \-  of  1  =  ^. 

3x(l)  =  (2)  ?  of  |  =  3x^\  =  ^  =  f,  result. 

3x2 
Process  Shortened  :      ?  of  |  = =  f ,  result. 

7x3 

By  the  short  process^  we  express  all  the  numbers  as  fractions, 
then  multiply  all  numerators  together  for  a  new  numerator, 
and  all  denominators  together  for  a  new  denominator.  Employ 
cancellation. 

4.  Find  i  of  I  of  f . 

Process  :  ^  of  |  of  f  =  l^i^^  =  j,  result. 

;2x^x^ 
3 

3 

Process:    71x^3^  =  ??^=/^,  result. 

2x^^ 
5 


6.  Multiply  I  of  8  by  J  of  6. 


Process  ;  f  ot  8 x  ^  of  6  =  ^^^^^^^  =  12,  result. 


2 

^x3 


Note. — In  the  results  all  improper  fractions  should  be  reduced  to  in- 
tegers or  mixed  numbers,  and  all  fractional  parts  should  be  reduced  to 
their  lowest  terms. 


1. 

7xtt 

7. 

2. 

AX68 

8. 

3. 

AofK 

9. 

If^ 

f  of  \%  of  150 

10. 

5. 

17^X^1^X280 

11, 

6. 

Aof2|Xifofy 

12. 

60  ADVANCED   ARITHMETIC. 

EXEKCISE  XXVII. 

Find  the  value  ^f — 

AX-rVXlfof  18| 
SxTAX-iVe^xOG 
126|X4iXTVTof  683 
16fx80|Xlif 
127AX45|X450 

ffx-A^x«fx|i 

43,  Division  of  Fractions. — The  easiest  way  to  divide 
in  fractions  is  to  invert  the  divisor  and  proceed  as  in  multiplication 
of  fractions.     To  explain,  we  know  that — 

1  apple  contains  |  apple  2  times. 
1  apple  contains  ^  apple  8  times. 
1  apple  contains  \  ai)ple  4  times. 
1  apple  contains  f  apple  2  times. 
1  apple  contains  ^  apple  6  times. 
1  apple  contains  f  apple  8  times,  etc. 

Now,  let  us  arrange  these  expressions  in  an  abstract  form : 

l-^i=f,or2. 
l-^i=f,  or8. 
l-^i=f,  or4. 
1-^1=1,  or  2. 
l--i=f,  or6. 
l-f=:|,  or8. 

Look  at  each  divisor  and  the  first  form  of  the  corresponding 
quotient,  and  you  will  see  that,  in  each  case,  the  quotient  is  the 
divisor  inverted.  This  is  an  illustration  of  the  general  truth  :  In- 
verting a  divisor  shows  how  many  times  that  divisor  is  contained  in 
one.  Commit  this  to  memory.  With  this  truth  in  mind,  let  us 
study  the  following  examples : 


ADVANCED   ARITHMETIC.  61 

EXAMPIiBS. 

1.  Divide  14  by  J. 

7  8 

Process  Complete  :   (1)     1-h  — =  — .     (Above  truth.) 

8  7 

7  2      8 

14x(f)  =  (2)  14+ -=;^x-  =  16,  result. 

8  fl 

7      2      8 
Process  Shortened :        14-f  -  =  ^^  x  -  =  16,  result. 

2.  |-^J=(      )? 

7     8 
Process  Complete  :    ( 1 )   1  +  —  =  — . 

2 

3  3     7     3^6 

jof(l)  =  (2)        --.-  =  - of  ^  =  -,  result. 

2 
3     7     3^6 

Process  Shortened :  ~-*-~  =  tX^  =  ~>  result. 


^.  lf-^8  =  (     )? 

Process  Complete  :    (1)  l-«-8  = 


4     8^77 
1 


3 

12  12       ;;2     1    3 

r7of(l)  =  (2)    -^8  =  j^of^=-,  result. 


12       ;;2   1    3 

Process  Shortened :     —  -5-8  =  —  X  —  = — ,  result. 

2 
Note. — The  first  step  in  the  complete  process  gives  us  the  divisor  in- 
verted ;  this  may  be  done  mentally.  The  ^'Shortened  Process"  is  simply 
the  last  step  in  the  Complete  Process,  and  may  be  expressed  thus:  Invert 
the  divisor  and  proceed  as  in  multiplication  of  fractions.  Use  the  shorter 
process  in  practice. 

4.  Divide  7i  by  8^. 

3 

■JR  q  Q 

Process  :  7i  -»-  31  =  —  x  —  =  -,  or  2^,  result. 
2/0     4 
2 


62  ADVANCED    ARITHMETIC. 

Note. — tt"  ^^  a  complex  fraction,  and  means,  like  any  other  fraction, 

that  its  numerator,  12^,  is  to  be  divided  by  its  denominator,  IJ.    Then 

the  process  is  easy : 

12t  61     2     122 

Process:  -— =  12^+11=  — x-  =  -—,  or  8^^,  result. 
1^  5      d      15 

When  the  terms  of  the  dividend  are  divisible  by  the  respec- 
tive terms  of  the  divisor,  division  of  fractions  may  be  i3er- 
formed  in  that  way. 


^.  t¥2-H=(    )^ 


88     11     884-11      8 

Process  :  7—  -»-  —  =  — =  — ,  result. 

132    12    132  +  12    11' 


EXERCISE  XXVIII. 


Find  the  value 

of- 

1.  -H-^SO 
4..  18I-M50 

5. 
6. 
7. 
8. 

18A-38i 

9     ^^ 
"^^    n     8A 

44.  Greatest  Common  Divisor  of  Common 
Fractions. —  The  Greatest  Common  Divisor  of  two  or  more 
fractions  is  the  largest  fraction  that  is  contained  in  each  of 
them  an  integral  number  of  times. 

One  fraction  is  a  divisor  of  another,  when  the  numerator  of 

the  first  is  a  divisor  of  the  numerator  of  the  second,  and  the 

denominator  of  the  first  is  a  multiple  of  the  denominator  of 

the  second.     Thus, 

/y  is  a  divisor  of  |. 

This  is  plain  when,  according  to  the  process  of  division  of  fractions, 
the  divisor  is  inverted. 


ADVANCED    ARITHMETIC.  63 

The  numerator  4  of  the  divisor  is  a  divisor  of  the  numerator  8  of  the 
dividend,  and  will  disappear;  the  denominator  27  of  the  divisor  is  a 
multiple  of  the  denominator  9  of  the  dividend,  and  the  9  will  disappear. 
When  the  4  and  9  both  disappear,  the  quotient  will  be  an  integer,  and 
jV  is  a  divisor  of  f . 

Principle  :  The  Greatest  Common  Divisor  of  two  or  more  frac- 
tions is  that  fraction  whose  numerator  is  the  G.  C.  D.  of  the  nu- 
merators of  the  several  fractions  and  whose  denominator  is  the 
L.  C.  M.  of  their  denominators. 

EXAMPLES. 

1.  Find  the  G.  C.  D.  of  J|,  H,  ^\,  and  |. 

(1)  The  G.C.D.  of  15,  25,  5,  5  =  5. 

(2)  The  L.  C.  M.  of  32,  64,  16,  8  =  64. 
.*.  -iiis,  the  required  G.  0.  D. 

2.  Find  the  G.  C.  D.  of  14^  and  95f . 

(1)  Expressed  as  fractions,  14rV  and  95f  =  V/  and  ^^. 

(2)  The  G.O.D.  of  175  and  763  =  7. 

(3)  The  L.  0.  M.  of  8  and  12  =  24. 
.'.  /j  is  the  required  G.  0.  D. 

EXERCISE  XXIX. 

Find  the  G.  C.  D.  of— 
1.  I,  A,  and  ||. 
^.  AV,  If,  if,  and  If. 
S-  «,  if,  f ,  A%,  and  T% 

4.  5A,  12f ,  and  4^. 

5.  9|,  5|,  16|,  and  8J. 

6.  f ,  3f ,  T^,  H,  and  5^. 

r.  8^1^,  9f ,  -U,  8A,  and  23-^. 
^.  nh  Ih  10|,  25i,  and  16f . 
9.  10/^,  17|,  12^\,  26t,  and  18^. 
iO.  3i,  2-,\,  12i,  9l,and8J. 


64  ADVANCED   ARITHMETIC. 

45.  Least  Common  Multiple  of  Common 
Fractions. —  The  Least  Common  Multiple  of  two  or  more 
fractions  is  the  smallest  number  that  will  contain  each  of 
them  an  integral  number  of  times. 

One  fraction  is  a  multiple  of  another,  when  the  numerator 

of  the  first  is  a  multiple  of  the  numerator  of  the  second,  and 

the  denominator  of  the  first  is  a  divisor  of  the  denominator  of 

the  second.     Thus, 

f  is  a  multiple  of  ^. 

Indicate  the  process  of  division  by  inverting  the  divisor.    Thus, 

The  8  is  a  multiple  of  4,  and  the  4  will  disappear;  the  9  is  a  divisor  of 
27,  therefore  the  9  will  disappear,  and  the  quotient  will  be  integral. 

Principle  :  The  Least  Common  Multiple  of  two  or  more  fractions 
is  that  fraction  whose  numerator  is  the  L.  CM.  of  the  numerators 
of  the  several  fractions^  and  whose  denominator  is  the  G.  C.  D.  of 
their  denominators. 

EXAMPIiES. 

1.  Find  the  L.  C.  M.  of  f,  A,  and  fi. 

(1)  The  L.  C.  M.  of  3,  9,  21  =  63. 

(2)  The  G.  0.  D.  of  4, 16,  32  =  4. 

.*.  ^  or  15|  is  the  required  L.  0.  M. 

2.  Find  the  L.  C.  M.  of  14f ,  9^\,  16|,  and  25. 

(1)  The  numbers  expressed  as  fractions  are^  ?^,  Yr",  ^^-,  \K 

(2)  The  L.  C.  M.  of  100,  100,  50,  25  =  100. 

(3)  The  G.  C.  D.  of  7,  11,  3,  1  =  1. 

.'.  ^^  or  100  is  the  required  L.  0.  M. 

EXERCISE  XXX. 
Find  the  L.  C.  M.  of— 

1'   A,  A^,  and  ii.  6.  ,V,  ^%,  4^^,  12f ,  and  4^V. 

^.    I,  A,  ii,  and  \.  7.  18f ,  15|,  19|,  and  ^. 

S-   h  A,  A,  and  H.  8.  87i,  Ih  SJ,  10|,  and  18f . 

4.    83J,  121,  6|,  and  7|-.  9.  22^^,  88|,  621,  5f ,  and  166J. 

^.    IH,  Hh  m.  -h.  and  -^S-  10.  142|,  55|,  41|,  and  888^. 


ADVANCED    ARITHMETIC.  65 

J.    DECIMAL  FRACTIONS. 

46.  Definitions. — A  Decimal  Fraction  has  for  its 

expressed  denominator  1  with  O's  annexed.     Thus : 

10>    T0"0>    TO^O^J    To  0  0  0*  I 

A  decimal  fraction  may  be  written  without  using  figures  for 
its  denominator.     Thus : 

Tfo  =  .05. 

AW^=.2045. 

By  this  method  of  writing  decimal  fractions,  (1)  the  numer- 
ator is  written  just  as  an  integer;  (2)  a  period,  called  a  Deci- 
mal Point,  is  put  on  the  left  of  as  many  places  in  the 
numerator  as  there  are  O's  in  the  expressed  denominator.  If 
there  are  not  so  many  places  in  the  numerator  as  are  needed, 
O's  are  prefixed  to  the  numerator  to  make  the  required  number 
of  places. 
A  decimal  fraction  written  in  this  way  is  called  a  Decimal. 
The  denominator  of  a  decimal  — 

of  one  place  is  10;    .7  is  read  y\; 

of  two  places  is  100 ;    .46  is  read  -^^ ; 

of  three  places  is  1000 ;    .807  is  read  ^^^. 

of  four  places  is  10000 ;    .0453  is  read  tUoq  y   ^^^  ^^  ^"• 

The  decimal  places  are  named  from  the  decimal  point  to  the 
right,  as  follows : 

The  first  place,  tenths  ; 

the  second  place,  hundredths ; 

the  third  place,  thousandtlis  ; 

the  fourth  place,  ten-thousandths  ; 

the  fifth  place,  hundred-thousandths ;  and  so  on. 


66 


ADVANCED    ARITHMETIC. 


The  decimal  orders  are  numbered  from  tenths  to  the  right 
as  the  integral  orders  are  numbered  from  units  to  the  left. 
Thus, 

Integral  Places.  Decimal  Places. 


'     . 

~~"^  ^~ 

— ^                -^ 

~"~~^ 

^"™^ 

CD 

CD          a. 

03           tr 

(/ 

tr 

to            cr 

{/ 

(» 

'O 

'd       t: 

t: 

c 

-»^ 

,C 

rC 

X 

X 

^ 

rj 

fl        c 

O)             0. 

■^ 

•4^               -*-: 

-tj 

■»3 

od 

§          K 

»- 

+= 

5               c 

'C 

t: 

t: 

•d 

DO 

So          CO         73 

a 

a>        c 

p 

c 

p 

D          :: 

c 

4J 

t- 

d 

« 

03 

o 

O          C 

'O            V 

tf 

O) 

M 

,c!         ^ 

^ 

c 

3 

z 

S 

43 

+3         +i 

3            C 

c 

O 

■i 

1 

X 

4^ 

,c 

5 

S 

^ 

p 

'd 

'd 

O)            (D 

c 

4- 

»H 

p 

T? 

A 

13 

* 

^^ 

• 

• 

6 

^ 

it     ^ 

)    1 

5     ^ 

r     ^ 

>       .       8      8      9       1      7 

. 

0 

. 

rC 

^ 

M 

a 

4J 

-t-3 

X 

X 

-U 

■4J 

t3 

•73 

.^ 

a 

M              'O              'U              ->^ 

-u 

CD 

»c 

^ 

CC 

i-H            CVJ            CO            Tt 

lO 

Integral  Orders. 


Decimal  Orders. 


The  law  of  position  (or  order)  in  decimals  is  the  same  as 
that  in  integers. 

Law  :  Ten  units  of  any  one  order  make  one  unit  of  the  next  higher 
order. 

The  denominator  of  a  decimal  is  indicated  by  the  name  of 
its  right-hand  place.     Thus,  in 

.0246, 

the  name  of  the  right-hand  place  is  ten-thousandths;  the  de- 
nominator is  ten  thousand. 

When  a  decimal  is  annexed  to  an  integer  the  expression  is 
called  a  Mixed  Decimal.     Thus: 


5.04;  and  is  read,  5^^^. 


ADVANCED   ARITHMETIC. 


67 


A  common  fraction  may  be  annexed  to  a  decimal ;   such  an 
expression  is  called  a  Coiiij)lex  Decimal,     Thus : 


m 

100* 


.12^;  and  is  read, 
6.87i;  and  is  read,  5 


37i 
100* 


In  an  expression  containing  two  or  more  decimal  points  all 

decimal  points  may  be  omitted  except  the  one  farthest  to  the 

left.     Thus : 

59.6.4=59.64. 


Explanation :  59.6.4=  59.6y\: 


59f: 


59 


n. 


59t%%=  59.64. 


There  is  no  such  expression  in  our  system  of  notation  as  a 
common  fraction  with  a  decimal  point  on  the  left  of  it.     Thus, 

•i- 

Explanation :  J  without  the  point  is  I  of  one  integral  unit.  .0^  is  J  of 
of  one-tenth,  which  is  the  first  decimal  order.  Therefore  there  is  no 
place  in  our  system  for  such  an  expression  as  .J. 


EXERCISE  XXXI. 


Read  : 

1, 

.6 

6. 

.049 

2. 

.25 

7. 

51.846 

S, 

.08 

8. 

.00.01 

4. 

.73 

9. 

7.0884 

6,  12.5 

10. 

.759.63 

Write  as 

decimals : 

16. 

•A 

21. 

tW^ 

17. 

3i 

22. 

? 

10 

1000 

18. 

10 

23. 

8J 
1000 

19. 

AV 

21,. 

IST^ftr 

20. 

37i 
*100 

25. 

^^A 

11.  100.00001 

12.  .04567 

13.  17.874 
U.        .06511 
15.  148.00ff 


26.   59_A_ 

10000 
27.      lOOyoljW^ 


28.    11 


iiiiixlr 


68  ADVANCED   ARITHMETIC. 

47.    Reducing  Decimals  to  Higlier  Terms. — 

If  a  0  is  annexed  to  a  decimal,  the  numerator  is  multij^lied  by 
10;  but  since  another  decimal  place  is  thus  added,  the  denom- 
inator is  also  multiplied  by  10,  and  the  value  of  the  fraction  is 
not  changed.     Thus: 

.5  =  . 50=. 500=. 5000,  etc. 
That  is,  A-t^«o=i^^=tWA,  etc. 

EXAMFIiE. 

Reduce  .125  to  lOOOOOths. 

Process  :  .125=  .12500,  result. 

Explanation :  Annex  O's  to  the  decimal  till  the  number  of 
decimal  places  is  equal  to  the  number  of  O's  in  the  required 
denominator. 

EXERCISE  XXXII. 


Reduce : 

1.    .7  to  lOOOths. 

6. 

5.1  to  lOOths. 

2.    .506  to  lOOOOths. 

7. 

7o05  to  lOOOths. 

3.    .94  to  lOOOths. 

8. 

56.084  to  lOOOOths.. 

4.    .08  to  lOOOOths. 

9. 

15*  to  lOOOths. 

5.    .004  to  lOOOOths. 

10. 

14.506  to  lOOOOOths 

48.   Reducing  Decimals  to  Lower  Terms. — If 

there  are  O's  at  the  right  of  a  decimal,  they  may  be  dropped 
without  changing  the  value  of  the  decimal.     Thus : 

.2300=. 28;  that  is, 


tWA^tW 

KXAMPIiES. 

1. 

Reduce  . 

.750  to  lOOths. 

Process 

:  .750  =.75,  result. 

2. 

Reduce 

.24000  to  lOOOths. 

Process 

;  .24000  =.240,  result. 

'  simply  place  the  decimal  point  after  the  15  and  add  the  O's. 


ADVANCED   ARITHMETIC, 


69 


EXERCISE  XXXIII. 


Reduce  • 


1. 

.200  to  lOths. 

6. 

5.0000  to  an  integer. 

2, 

.3400  to  lOOths. 

7. 

7.5000  to  lOths. 

3. 

.10500  to  lOOOths. 

8- 

15.0700  to  lOOths. 

4. 

.50000  to  lOOths. 

9. 

70.0700  to  lOOths. 

5. 

.94000  to  lOOths. 

10. 

.005000  to  lOOOths. 

49.  Addition  of  Decimals. — Write  the  numbers  to 
be  added  so  that  the  decimal  points  will  form  a  column.  Add 
as  in  integers,  placing  the  decimal  point  in  the  result  beneath  the 
column  of  decimal  points  above. 


EXAMPIiS. 


1.  Add  .501,  3.45,  .125,  7.0034,  18.3024. 


Long  Way. 

.5010 

3.4500 

.1250 

7.0034 

18.3024 


Short  Way, 

.501 
3.45 
.125 

7.0034 
18.3024 


Explanation  :  In  the  long  way,  all 
decimals  are  reduced  to  the  highest 
denomination,  lOOOOths.  The  de- 
nomination of  the  sum  is  lOOOOths. 
Why  ? 
29.3818,  result.  In  the  short  way,  the  decimals  are      29.3818,  result. 

not  reduced    to    higher  denomina- 
tions.    The  O's  on  the  right  do  not  affect  the  result.    Use  the  short  way 
in  practice. 

EXERCISE  XXXIV. 
Add: 

1.  .52,  .936,  1.63,  .0075. 

2.  .428,  .506,  .362,  .521,  .736, 

3.  .002,  5.3043,  7.5555,  81.8008, 

4.  12.342,  18.9813,  85.753,  .00059. 

5.  8.756,  .00045,  .04621,  .989898,  .OIOIOL 

6.  $25.30,    $846.27,    $954.63,    $1205.84,   $746.98,  $1375.30, 

$3.85,  $840.75,  $1684.35. 

Note. — When  an  expression  of  $'s  has  only  two  decimal  places,  call 
the  decimal  part  cents  ;  if  there  are  more  than  two  decimal  places,  read 
the  decimal  part  as  a  fraction  of  a  dollar. 


70  ADVANCED   ARITHMETIC. 

7.  $44.56,   $752.28,    $984.81,    $1200.50,   $845,95,   $7400.20, 

$947.25,  $75.75. 

8.  $56,845,  $824,955,  $5476.875,   $1260,  $95,625,  $4000.50, 

$1862.05. 

9.  586.084,   7495.6084,   845.9567,    1111.1111,   9988.7766, 

1200.0001,  .99864. 
10.  .04654,  .90083,  .12456,  .00665,  .77777,  .854821,  .750864, 
.999999. 

50.  Subtraction  of  Decimals. — Write  the  numbers 
so  that  the  decimal  point  in  the  subtrahend  will  be  beneath 
the  decimal  point  in  the  minuend.  Subtract  as  in  integers,  plac- 
ing the  decimal  point  in  the  result  beneath  the  decimal  points  above. 

EXAMPIiES. 

1.  From  5.47  take  2.688. 

Process :  Explanation :   Reduce  the  minuend  to  the  de- 

5.470  nomination  of  the  subtrahend  by  annexing  0  to 

2-638  it.    The  denomination  of  the  result  is  lOOOths, 

2.832,  result.  Why  ? 

2.  From  .54652  take  .869. 

Long  Way.  Explanation :  In  the  long  way,  the    ^^^^^  ^^^^^ 

54652  subtrahend  is  reduced    to  the  de-         54652 

.36900  nomination  of    the  minuend.     The        '359 

J77T2,  result.     ^^^T^Tr.""!   the    result    is        ^752,  result. 
lOOOOOths.    Why  ?    In  the  short  way, 
the  subtrahend  is  not  reduced  to  higher  denominations.    The  O's  on  the 
right  do  not  affect  the  result.    Use  the  short  way  in  practice. 

EXERCISE  XXXV. 


^ubt\ 
1. 

ract  : 
.2464  from  .65. 

6. 

8.427  from  8.245. 

2. 

.88  from  5.001. 

7. 

.0847  from  .5087. 

3. 

.428  from  .6. 

8. 

.0042  from  .042. 

4. 

.5748  from  .8046. 

9. 

$1.20  from  $7.05. 

5. 

1.5644  from  5. 

10. 

$246.75  from  $4000. 

ADVANCED    ARITHMETIC, 


71 


51.  Multiplication  of  Decimals. — Multiply  as  in 
integers,  neglecting  O^s  that  may  be  on  the  left  of  the  multiplicand 
and  multiplier,  and  place  the  decimal  point  in  the  result  so  that 
there  will  be  as  many  decimal  places  in  the  product  as  there  are  in 
both  multiplicand  and  multiplier. 


BXAMPIilSS. 


1.  Multiply  .852  by  .35. 


Process, 


.8  5  2 
.3  5 

4260 
2556 

.29820,  result. 


Explanation :  The  multiplicand  is  thou- 
sandths ;  the  multiplier  is  hundredths. 
Hundredths  times  thousandths  gives  hun- 
dred-thousandths. Show  this  by  use  of 
common  fractions.  Hundred-thousandths 
is  expressed  by  a  decimal  of  5  places.  Af- 
ter placing  the  decimal  point,  cancel  the 
0*s  on  the  right;  for  the  product  should 
be  expressed  in  its  lower  terms. 


Multiply  .0037  by  2.5. 


Process. 


.00  3  7 
2.5 

185 

74 

.00925,  result. 


Explanation :  For  the  purpose  of  the 
multiplication,  the  multiplicand  is  37 
(neglecting  the  O's  on  the  left). 

The  product  after  multiplying  contains 
but  three  figures ;  but  there  must  be  five 
decimal  places.  Why  ?  Therefore  two 
O's  are  prefixed  and  the  decimal  point 
is  placed  to  the  left. 


3.  Multiply  246  by  3.07. 
Process. 


246 

3.0  7 


1722 
738 

7  5  5.2  2 


Note. — Have  pupil  explain. 


72  ADVANCED   ARITHMETIC. 


EXERCISE 

XXXVI. 

Multiply  — 

1.   .75  by  .23. 

7.   75.003  by  5.006. 

2.    .428  by  .501. 

8.    1000.001  by  100.0001. 

S.  4.763  by  .084. 

9.    1111.22  by  5.060701. 

4.    .00073  by  .0059. 

10.   $420.50  by  .37. 

5.    .5064  by  .00001. 

11,   $4860.05  by  246. 

6.   56.043  by  .059. 

12,    $840.56  by  10.05. 

52.  Division  of  Decimals. — Divide  as  in  integers, 
neglecting  O^s  that  may  be  on  the  left  of  the  dividend  or  divisor. 

Since  the  dividend  is  equal  to  the  product  of  the  divisor  and 
quotient  (Art.  14,  Prin.  3),  the  dividend  v/ill  have  as  many 
decimal  places  as  both  divisor  and  quotient.  From  this,  the 
following  conclusions  are  evident : 

(1)  The  dividend  must  not  have  fewer  decimal  places  than  the 
divisor,     (Commit.) 

Note. — If  the  dividend  has  not  as  many  decimal  places  as  the  divisor, 
enough  O's  to  make  them  equal  should  be  annexed  to  the  dividend  before 
dividing. 

EXAMPIiE. 

Prepare  to  divide  .15  by  .0025. 

Explanation:  Two  O's  must  be  an- 
Form.  nexed  to  the  dividend,  because   there 

— are  four  decimal  places  in  the  divisor 

.0025;  .1500  and  only  two  decimal  places  in  the  div- 

idend. 

(2)  7/,  after  dividing^  the  number  of  decimal  places  used  in  the 
dividend  equals  the  number  in  the  divisor y  the  quotient  is  an  intC' 
ger,     (Commit.) 


ADVANCED    ARITHMETIC. 


73 


Divide  .15  by  .0025. 

Process. 

60,  result.  ' 

.0025). 1500 
150 


£XAMPI/E. 


Note. — In  the  actual  process  of  divi- 
sion, the  two  O's  on  the  left  of  the  divi- 
sor are  not  used ;  the  divisor  is  25. 


■  (8)  IJy  after  dividing^  the  number  of  decimal  'places  used  in  the 
dividend  exceeds  the  number  of  decimal  places  in  the  divisor,  the 
number  of  decimal  places  in  the  quotient  must  equal  that  excess. 
(Commit.) 


EXAIfPIi^S. 


1.  Divide  .98745  by  .29. 
3.4  0  5,  result. 


.2  9).9  8  7  45 

87 

117 
116 


145 
145 


Explanation  :  There  are  Jive  deci- 
mal places  in  the  dividend  and  two  in 
the  divisor.  The  excess  is  three. 
Therefore,  the  quotient  must  con- 
tain three  decimal  places. 


K  Divide  .002868  by  .239 

Process. 

.012,  result 


.239). 002868 


478 
478 


Exjilanation  :  After  dividing  the  2868 
(neglecting  the  O's  on  the  left)  by  239, 
the  quotient  is  12;  but  tjje  excess  of 
decimal  places  in  the  dividend  over 
those  in  the  divisor  is  three.  There- 
fore, the  quotient  must  contain  three 
decimal  places. 


If  there  is  a  remainder  after  using  the  last  figure  of  the  divi- 
dend, O^s  may  be  annexed  to  the  dividend  and  the  division  continued 
at  pleasure. 


74 


ADVANCED    ARITHMETIC. 


EXAMPIjE. 


Divide  2.09  by  .017. 
Process. 


12  2.941  17  +  ,  result. 


.0  17)2.09  000  00  0 
17 


39 
34 


50 
34 


160 
153 


70 
68 


20 
17 


30 
17 

130 
119 

11 


Explanation  :  The  plus 
sign  after  the  quotient 
shows  that  there  is  a  re- 
mainder, and  that  the  di- 
vision could  be  continued 
farther,  if  desired.  After 
dividing  as  far  as  desired, 
locate  the  decimal  point 
according  to  Rule  3. 


EXERCISE  XXXVII. 


ivii 

ie  — 

1. 

.125  by  .5. 

7. 

.025  by  5. 

2. 

.5  by  1.28. 

8. 

.025  by  .5. 

3. 

.12  by  .0016. 

9. 

1000  by  .001. 

4. 

.12  by  16. 

10. 

.5484888  by  .0067. 

5, 

12  by  .16. 

11. 

$4826.422  by  $96.1. 

6. 

.001  by  1000. 

12. 

104.8576  by  .001024. 

53.  Reducing*  Common  Fractions  to  Deci- 
mals.—  Since  a  common  fraction  is  an  expression  of  division, 
if  the  division  be  performed,  the  result  will  be  an  integer  or  a 
decimal.  In  dividing,  observe  the  rules  for  division  of  deci- 
mals. 


ADVANCED    ARITHMETIC.  75 

BXAMFIjSS. 

1.  Reduce  .7^  to  a  decimal. 

Process. 

.18  75 
16)3.0000 

^  ^  Exjilanation  :  By  reducing  the  i\  to 

14  0  a  decimal  and  annexing  it  to  the  .7,  the 

^  ^  ^  result  is  .7.1875.     But  all  decimal  points 

12  0  may  be  omitted  except  the  one  on  the 

left. 


112 


80 
80^ 

,7.1875  =  .71875,  result. 

S.  Reduce /^  to  a  decimal.     (Continued  4  places.) 

Explanation  :  It  will  be  observed  that 

rrocess.  ^j^j^  division  w^ill  never  be  complete  ;  for 

.2  12  1  +  ,  result.  7  ^as  the  figure  in  the  dividend  at  the 

3  3)7.0  0  0  0  beginning.    It  was  the  remainder  after 

^  ^  the  2d  division,  and  also  after  the  4th 

4  0  division.    In  such  a  process,  the  divi- 

33  sion  will  never  be  complete,  if  the  de- 

7  0  nominator  has  any  other  factor  than  2 

8  6  or  5.  These  are  factors  of  10,  and  by 
4  0  adding  O's  to  the  numerator  we  multi- 
33  ply  by  10.    The  numerator  can  thus  be 

7  made  to  contain  any  number  of  2's  or 

5's,  but  no  other  factor. 

In  such  a  decimal  as  is  obtained  in  example  2  above,  one  or 
more  figures  continue  to  recur  in  a  certain  order.  Such  deci- 
mals are  called  Recurring'  or  Circiilating"  Decimals, 
or  simply  Circulates.  The  part  that  recurs  is  called  the 
Repetend,  and  is  marked  by  placing  a  dot  over  the  first 
and  last  figures  of  the  repetend.     Thus, 

5,  .2i84. 


76  ADVANCED    ARITHMETIC. 

The  first,  if  written  out,  would  be  — 

.5555555 to  infinity. 

Note. — "To  infinity ^^  means  that  the  expression  would  never  termi- 
nate, but  would  continue  forever  as  indicated. 

The  second,  if  written  out,  would  be  — 

.2134134134134 to  infinity. 

3.  Reduce  ^  to  a  circulating  decimal. 
Process. 

A  3  2,  result. 


37)16.000  Explanation:    After  three  divisions 

^  ^  ^  the  remainder  is  16,  which  is  the  same 

12  0  as  at  the  start.    Therefore  the  repetend 

111  consists  of  3  figures. 


90 
74 

16 

educe  -^ 

to  a  circulate. 

Process. 

.3i8, 

result.          E^P^ 

Explanation :    The  same  remainder 

9  9)7  0  0  0  recurs  after  the  first  and   third  divi- 

QQ  sions;  therefore  the  second  and  third 

.  ^  divisions  will  be  repeated   as  long  as 

2  2  the  division  is  continued,  and  the  repe- 

TTT^  tend  is  18,  commencing  with  the  second 

1  7  g  decimal  place. 

4 

EXERCISE    XXXVIII. 

Reduce  to  decimals,  obtaining  complete  results  : 

1.  i  I,.    12|f  7.    73/J-^ 

2.  \\  5.   48H  8.    .5f 

3.  II  6.   3.7^V  9.    .064* 
Reduce  to  decimals,  obtaining  results  true  to  four  places  : 

10.  A  13.    H  16.    .07^^^ 

11.  -H  U.    12H  17.    84.9^iA 

12.  -jV  15.    8.4f  18.    741.07Ji 


ADVANCED   ARITHMETIC.  77 

Reduce  to  circulating  decimals  : 

19.  i  22,    t\  ^5-    12iH 

20.  -ii  23.    iU  ^^.   rU 

21.  \  2Jt..   |§  27.    15|f 

54.  Reducing'  Decimals  to  Commoii  Frac- 
tions,—  Express  the  denominator,  which  is  1  with  as  many 
O's  annexed  as  there  are  decimal  places.  Then  reduce  the  re- 
sult to  its  simplest  form. 

EXAMPLES. 

1.  Reduce  .15  to  a  common  fraction. 

Process  ;  .15  =  ii(^=  2^,  result. 

2.  Reduce  .0036  to  a  common  fraction. 

Process  :  .0036  =  iuVots  =  a  ^gu »  result. 

S.   Reduce  5.125  to  a  mixed  number. 

Process:  .125  -=^js%%  =  h 

.'.  5i  is  the  required  result. 

Note. — Reduce  the  decimal  part  to  a  common  fraction  and  annex  it 
to  the  integer. 

4.  Reduce  .12^  to  a  common  fraction. 

Process:  .12^  =  ^=12J^100=«2^XTh  =  i. 

5.  Reduce  .6  to  a  common  fraction. 

Process  :    (1)  .6  =  .666666 to  infinity. 

10  X  (1)  =  (2)  10  X  .6  =  6.666666 to  infinity. 

(2)-(l)  =  (3)  9x.6  =  6. 
iof(3)  =  (4)  .6  =  1  =  1,  result. 

Note. —  Subtracting  the  repetend  from  10  times  the  repetend  leaves  9 
times  the  repetend.  Subtracting  .666666  to  infinity  from  6.666666  to  in- 
finity leaves  6.  Therefore  9  times  the  repetend  equals  6,  and  the  repe- 
tend equals  |. 


78  ADVANCED   ARITHMETIC. 

6.  Reduce  .507  to  a  common  fraction. 

Process  :    (1)  .507  =  .507507507 to  infinity. 

1000x(l)  =  (2)  1000 X. 507  =  507.507507 to  infinity. 

(2)-(l)  =  (3)  999 X. 507  =  507. 
^of  (3)  =  (4)  507  =  m=J||,  result. 

Principle  :  Any  repetend  may  he  expressed  as  a  common  frac- 
tion hy  writing  the  repetend  for  the  numerator  and  as  many  9^s  for 
the  denominator  as  there  are  figures  in  the  repetend. 

7.  Reduce  17.54  to  a  mixed  number. 

Process  :     .54  =  f  f  =  j\ . 

.',    17.54  =  17i\,  result. 

8.  Reduce  .25108  to  a  common  fraction. 

Process  :    (1)  .25i08  =  .25iif  =  .25iVi 

(2)  .25iVr  =  ^  =  W  =  iViVo  =  Wd  ,  result. 

Note. — In  processes  where  circulating  decimals  are  involved,  reduce 
the  circulate  to  a  common  fraction  before  beginning  the  process.  At 
the  conclusion,  the  result  may  be  expressed  as  a  circulate,  if  required. 

EXERCISE  XXXIX. 
Reduce  to  com,mon  fractions  or  mixed  numbers  : 


1. 

.875 

7. 

M 

13. 

.63 

2. 

.125 

8, 

.161 

u. 

.063 

3. 

.425 

9. 

.47S 

15. 

.248 

4. 

5.625 

10. 

.03711 

16. 

.572 

5, 

8.875 

11. 

2.88A 

17. 

.1252 

6. 

.025 

12, 

15.84,%V 

18. 

7.0i575 

ADVANCED    ARITHMETIC.  79 

II.    STUDY  OF  PROBLEMS. 
A.     THE  MEANS  OF  EXPRESSING  SOLUTIONS. 

1.     EQUATIONS. 

55.  Definitions. — A  statement,  in  which  the  sign  of 
equality  ( = )  is  used  ■  between  two  numerical  expressions  to 
show  that  they  are  equal,  is  an  Equation.  The  expression 
on  the  left  of  the  sign  is  called  the  first  or  left  member;  that  on 
the  right,  the  second  or  right  member.  When  a  member  of  an 
equation  is  separated  into  parts  by  either  of  the  signs,  plus 
(  +  )  or  minus  (  — ),  or  both,  these  parts  are  called  Terms  ; 
when  not  so  separated,  the  whole  member  is  called  a  term. 
See  Terms,  Article  21. 

EXAMPLES. 

1.  9-4+5=10. 

9,  4,  5, 10,  are  terms. 

2,  A's  money— $50=B'8  money. 

A's  money,  $50,  and  B's  money,  are  terms. 

S.  4  pecks  =  1  bushel. 

Each  member  is  a  term. 

4..  Area  of  a  surface  1  foot  long,  1  foot  wide=l  square  foot. 
Each  member  is  a  term. 

56.  Classification  as  to  Source. —  Some  equations 
express  constant  relations,  and  are  always  true;  other  equa- 
tions express  relations  depending  upon  conditions,  and  are 
therefore  true  for  particular  conditions  only. 

The  following  are  some  of  the  equations  which  are  always 
true: 

1.  6+4-7=3. 

2.  5X8=40. 

3.  56^8=7. 

4..  4  pecks  =  1  bushel. 

5.  Area  of  a  surface  1  foot  long,  1  foot  wide=l  square  foot. 

6,  8xB's  mouey+3xB's  money = 11 X B's  money. 


80  ADVANCED    ARITHMETIC. 

The  following  equations  are  true  for  particular  conditions 
only: 

-?.  Cost  of  5books=:$8.50. 

2.  -J  of  A's  age=B's  age +6  years. 

3.  Weight  of  1  load =2475  pounds. 

^.  Money  earned  by  7  men  in  2  days  =  $85. 

Note. — As  equations  of  the  first  class  are  always  true,  they  may  be 
used  in  the  solution  of  any  problem  where  such  use  is  of  service ;  but 
those  of  the  second  class  only  when  by  the  conditions  of  the  problem 
they  are  true. 

57.  Transformation. — An  equation  may  be  changed, 
or  transformed  from  a  given  to  a  required  form,  by  subjecting 
it  to  one  or  more  of  the  following  processes : 

1.  To  turn  an  equation  around. 

2.  To  transpose  the  terms  of  an  equation. 

3.  To  simplify  the  members  of  an  equation. 

4.  To  multiply  an  equation. 

5.  To  divide  an  equation. 

58.  To  Turn  an  Equation  Around, — To  turn  an 
equation  around  is  to  put  the  left  member  on  the  right  and 
the  right  member  on  the  left  of  the  sign  of  equality. 

Principle  :    Equals  are  equal  in  whatever  order  considered. 

It  follows  from  this  principle  that  any  equation  may  be 
turned  around  and  still  remain  a  true  equation. 

EXAMPIiES. 

1.  4  gills  =  l  pint;  then,  1  pint=4  gills. 

2.  Cost  of  5  hats  =  $15;  then,  $15=cost  of  5  hats. 

3.  $740— cost  price  =$560;  then,  $560=  $740— cost  price. 

Note. — This  process  is  so  easy  that  no  exercise  is  given  for  practice. 


ADVANCED   ARITHMETIC.  81 

59.  Transposition  of  Terms. — Transposing  a  term 
is  changing  it  from  one  member  of  an  equation  to  the  other. 

Note. — Transposition  is  looked  upon  by  many  pupils  as  beyond  the- 
sphere  of  arithmetic.  The"  following  exercise  is  given  with  the  aim  of 
suggesting  that  transposition  depends  upon  the  most  elementary  knowl- 
edge of  number  and  not  upon  some  difRcult  mathematical  principle. 

EXERCISE  XL. 

Answer  each  question,  then  tell  what  number  has  been  transposed  : 

i.  5+6=11;    then,  5=:ll-(   )? 

2.  5+6=11;    then,  6  =  ll-(   )? 

3.  9-5=4;    then,  9=4+(  )? 

4.  9=6+8;    then,  9- (   )=6? 

5.  7-4+9=12;    then,  7+9=12+(  )? 

6.  7-4+9  =  12;    then,  7-4=12-(  )? 

7.  7-4+9=12;    then,  9-4=  12-(  )? 

8.  5+4=12-3;   then,  5+4+ (  )  =  12? 

9.  14-8=4+2;   then,  14-8- (  )=2? 

Principles:  1.  If  equals  be  added  to  equals,  the  sums  will  be 
equal. 

2.  If  equals  be  subtracted  from  equals,  the  remainders  will  be 
equal, 

BXAMFIiES. 

1,  $75-$20=$25+$80;  transpose  $20. 

Process:     (1)  $75-$?0  =  $25+$3O. 

(2)  $?0  =  $2O. 
(l)  +  (2)  =  (3)  $75  =  $25+$30+$20. 

Explanation  :  If  (2)  be  added  to  (1),  the  sum  of  the  first 
members  will  be  $75  -  $20+ $20.  The  two  20's,  one  to  be  sub- 
tracted and  the  other  to  be  added,  balance  each  other  and 
may  be  canceled.  This  leaves  $75  for  the  first  member  of  (3). 
The  second  members  of  (1)  and  (2)  added  give  $25 +  $30 +$20. 


82  ADVANCED    ARITHMETIC. 

2,  A 's  money +$50= B's  money ;  transpose  A's  money. 

Process:    (1)  A^g  money  +  $50  =  B'8  money. 
(2)  A'g  money  =  A's  money. 
(l)-(2)  =  (3)  $50  =  B's  money -A's  money. 

Note. — Let  the  pupil  explain. 

3.  $40 -$80 =$60 -$50;  transpose  the  $50. 

Process  :    (1)  $40  -  $30  =  $60  -  $^0. 
(2)  $50  =  $^0. 
(l)+(2)  =  (3)  $40 -$30+ $50  =  $60. 

Note. — Let  the  pupil  explain. 

Each  of  these  exercises  and  examples  has  illustrated  one 
principle,  which  may  be  called  the  law  of  transposition. 

Law. — Any  term  of  one  member  of  an  equation  may  he  trans- 
posed to  the  other  member^  if  at  the  same  time  its  sign  is  changed  ^ 
the  plus  to  minus  or  the  minus  to  plus. 

4..  30—15+40=66  —  11;  transpose  11  and  15. 

Result:  30+40+11  =  66+15. 
Note. — The  process  should  be  mental. 

5.  51-12+5=25+19.     Transpose  the  12  and  the  19. 

Eesult ;  51 + 5  - 19  =  25 + 12. 

6.  100%  of  cost+$415=36i%  of  cost+$1050.  Transpose  so 
that  only  similar  terms  will  be  in  the  same  member. 

Result :  IW  of  cost  -  36^^  of  cost  =  $1050  -  $415. 

Note.— The  sign,  %,  is  read  per  cent;  and  it  means  hundredth  or  hun- 
dredthSf  according  as  it  is  used  with  one  or  more  than  one. 

EXERCISE  XLI. 

1.  Cost+$200=3Xcost+$80.  Transpose  ^'s  to  the  first 
member  and  cost  to  the  second. 

2.  Cost  of  15  hats=$70+cost  of  1  hat.  Leave  fs  only  in 
second  member. 


ADVANCED    ARITHMETIC.  8o 

3.  25—17+44—19=86  —  15+12.  Transpose  so  as  to  leave 
no  minus  signs. 

4.  75%  of  No. -200=40%  of  No. +200.  Transpose  so  that 
similar  terms  will  be  together,  per  cent  in  the  first  member. 

5.  i  of  the  tens+^  of  the  units =^  of  the  units.  Put  all  the 
units  in  the  second  member. 

6.  A's  money  — $600= i  of  A's  money— $200.  Transpose  so 
as  to  leave  no  minus  signs. 

7.  10  weeks'  wages  — $12=4  weeks'  wages+$24.  Transpose 
so  that  similar  terms  will  be  together,  #'s  in  the  second  mem- 
ber. 

8.  5Xmy  age +24  years =3  X  my  age +104  years.  Transpose 
so  that  similar  terms  will  be  together,  years  in  the  second 
member. 

60 o  To  Simplify  Members. — A  member  of  an  equa- 
tion is  a  numerical  expression.  By  transposition,  the  terms  of 
any  member  may  be  made  similar.  A  member  whose  terms 
are  similar  may  be  simplified  by  the  same  processes  that  are 
employed  to  simplify  other  numerical  expressions.  See  Arti- 
cle 21. 

EXAMPLE. 

1.  i  of  wt.+i  of  wt.+i  of  wt.=6  pounds+5  pounds+8 
pounds. 

Process:    (1)  i  of  wt.  +  ^  of  wt. +|  of  wt.  =6  pounds+5 
pounds  +  8  pounds. 
(1)  simplified  =  (2)  H  or  lTVxwt.  =  19  pounds. 

EXERCISE  XLII. 

Simplify  the  members  of  the  resulting  equations  in  Exercise  XLI. 

Note.— Before  attempting  to  simplify,  be  sure  that  similar  terms  are 
in  the  same  member. 

61.  To  Multiply  an  Equation. — An  equation  is 
multiplied  by  multiplying  each  of  its  members  by  the  same 
number. 


84  ADVANCED   ARITHMETIC. 

General  Principles  :  1.  Equals  multiplied  by  the  same  num- 
ber give  equal  iwoducts. 

2.  Multiplying  each  term  of  a  member  of  an  equation  by  the  same 
number  multiplies  the  member  by  that  number. 

Note. — In  addition  to  these  general  principles,  a  number  of  special 
principles  are  illustrated  in  the  following  examples.  They  will  be  given 
in  connection  with  their  respective  illustrations,  and  should  be  commit- 
ted to  memory. 

EXAKPIiES. 

i.  1  bushel  =4  pecks.     Multiply  by  5. 

Process:    (l)lbu.  =  4pk. 
*5x(l)  =  (2)  5bu.  =  20pk. 

2.  My  age =35  years.     Multiply  by  12. 

Process  :    (1)  My  age  =  35  yr. 
12  X  (1)  =  (2)  12  X  my  age  =  420  yr. 

3.  Cost  of  1  hat  =  $2.     Multiply  by  8.     (2  forms.) 

Process  :    (1)  Cost  of  1  hat  =  $2. 

8  X  (1)  =  (2)  8  X  cost  of  1  hat  =  $16 ;  or, 
8  X  (1)  =  (3)  Cost  of  8  hats  =  $16. 

Note. — Observe  that  either  equation  (2)  or  (3)  is  the  required  form, 
and  that  8  x  cost  of  1  hat  =  cost  of  8  hats. 

Principles:  1.  The  cost  of  any  number  of  articles  of  the  same 
kind  is  equal  to  that  number  of  times  the  cost  of  1  of  those  articles. 

4.  4+9-3+6=16.     Multiply  by  7. 

Process:    (1)4+9-3  +  6  =  16. 

7x(l)  =  (2)  28+63-21+42  =  112.     (Prin.2.) 

5.  .$50— A's  money=i  of  B's  money.     Multiply  by  8. 

Process:     (1)  $50- A's  money  =  J  of  B's  money. 
8x(l)  =  (2)  $400-8 xA's  money  =  4 x B's  money. 

*"5X(l)=-'(2) "  Is  read  5  times  equation  1  equals  equation  2. 


ADVANCED    ARITHMETIC.  85 

^.6X5=30.     Multiply  by  4.     (2  forms.) 

Process:   (1)  6x5  =  30. 
4x(l)  =  (2)  (4x6)x5  =  24x5  =  120;  or, 
4x(l)  =  (3)  6 X (4x5)  =  6x20  =  120.     (Prin.  1,  Art.  27.) 

7.  Area  of  a  surface  1  foot  long,  1  foot  wide  =  l  square  foot. 
Multiply  by  6.     (2  forms.) 

Process  :   (1)  Area  of  a  surface  1  ft.  1.,  1  ft.  w.  =  1  sq.  ft. 
6x(l)  =  (2)  Area  of  a  surface  6  ft.  1.,  1  ft.  w.  =  6  sq.  ft. ;  or, 
6  X  (1)  =  (3)  Area  of  a  surface  1  ft.  1. ,  6  f t.  w.  =  6  sq.  ft 

Note  1. — In  general,  length  means  the  longest  dimension  of  a  surface 
or  solid,  but  mathematically  it  simply  means  one  of  the  two  or  three  di- 
mensions, and  it  is  not  necessarily  the  longest. 

Note  2. — The  equation  in  the  above  example  expresses  a  certain  area 
— '^Area  of  a  surface  {of  certain  dimensions)  =  1  sq.  ft."  Any  process  which 
multiplies  the  area  of  this  surface,  multiplies  this  equation.  This  can 
be  done  by  multiplying  either  the  length  or  the  width,  but  not  both.  A 
surface  6  times  as  long  and  of  the  same  width  will  have  6  times  as  much 
area,  or  a  surface  of  the  same  length  but  6  times  as  wide  will  have  6 
times  as  much  area. 

2.  (1)  Multiplying  the  length  or  width  multiplies  the  area; 
therefore,  (2)  the  length  and  loidth  bear  to  the  area  the  same  nu- 
merical relation  as  that  of  factors  to  the  product. 

Note. — '* Numerical  relation"  means  relation  as  a  number  without 
considering  its  denomination  or  kind.  Two  numbers  expressing  feet 
(as,  4  feet  and  5  feet)  could  not  be  factors,  could  not  be  multiplied  to- 
gether, but  their  numerical  values  (4  and  5)  could  be  multiplied  together 
and  could  be  factors. 

Question  :  Does  Prin.  1,  Art.  27,  apply  to  example  7? 

8.  The  volume  of  a  solid  1  foot  long,  1  foot  wide,  and  1  foot 
thick  =  l  cubic  foot.     Multiply  by  12.     (3  forms.) 

Process  :   (1)  Vol.  of  a  solid  1  ft.  1. ,  1  ft.  w.,  1  ft.  th.  =  1  cu.  ft. 
12x(l)  =  (2)  Vol.  of  a  solid  12  ft.  1.,  1ft.  w.,  1ft.  th.  =  12  cu.  ft. ;  or, 
12x(l)  =  (3)  Vol.  of  a  solid  1ft.  1.,  12  ft.  w.,  1ft.  th.  =  12cu.  ft.;  or, 
12  X  (1)  =  (4)  Vol.  of  a  solid  1  ft.  1. ,  1  ft.  w. ,  12  ft.  th.  =  12  cu.  ft. 


86  ADVANCED    ARITHMETIC. 

3.  (1)  Multiplying  the  lengthy  widths  or  thickness  multiplies  the 
volume;  therefore^  (2)  the  lengthy  widths  and  thickness  bear  to  the 
volume  the  same  numerical  relation  as  that  of  factors  to  the  product. 

Question:  Does  Prin.  1,  Art.  27,  apply  to  example  8? 

9.  The  interest  on  $1  for  1  yr.  at  1%=$.01.  Multiply  by 
50.      (3  forms.) 

Process :   (1)  Int.  on  $1  for  1  yr.  at  1^  =  $.01. 
50x(l)  =  (2)  Int.  on  $50  for  1  yr.  at  1^  =  $.50;  or, 
50  X  (1)  =  (3)  Int.  on  $1  for  50  yr.  at  1%  =  $.50 ;  or, 
50x(l)  =  (4)  Int.  on  $1  for  1  yr.  at  50^  =  $.50. 

4..  (1)  Multiplying  the  principal,  time,  or  rate,  multijjlies  the  in- 
terest; therefore,  (2)  the  principal,  time.,  and  rate  hear  to  the  inter- 
est the  same  numerical  relation  as  that  of  factors  to  the  product. 

Question:  Does  Prin.  1,  Art.  27,  apply  to  example  9? 

10.  The  amount  of  wood  sawed  by  1  man  in  1  day=l  cord. 
Multiply  by  10.     (2  forms.) 

Process :  (1)  Amt.  sawed  by  1  man  in  1  da.  =  1  c. 
10  X (1)  =  (2)  Amt.  sawed  by  10  men  in  1  da.  =  10  c. ;  or, 
10  X  (1)  =  (3)  Amt.  sawed  by  1  man  in  10  da.  =  10  c. 

5.  (1)  Multiplying  the  number  of  working  units  or  the  time, 
multiplies  the  amount  of  work  done;  therefore,  (2)  the  working  force 
and  the  time  bear  to  the  work  done  the  same  numerical  relation  as 
that  of  factors  to  the  product. 

Question  :  Does  Prin.  1,  Art.  27,  apply  to  example  10? 

11.  8-^4  =  2.     Multiply  by  4.     (2  forms.) 

Process:   (1)  8-^4  =  2. 
•  4x(l)  =  (2)  (4x8)  +  4  =  32-h4  =  8;  or, 
4x(l)  =  (3)  8^(4^4)-8^1  =  8.     (Prin.  3,  Art.  27.) 


ADVANCED   ARITHMETIC.  87 

12.  The  length  of  a  surface  of  72  sq.  ft.,  8  ft.  wide =9  ft. 
Multiply  by  8.     (2  forms.) 

Process  :  (1)  Length  of  a  surface  of  72  sq.  ft. ,  8  ft.  w.  =  9  ft. 

8x(l)  =  (2)  Length  of  a  surface  of  576  sq.  ft.,  8  ft.  w.  =  72  ft. ;  or, 
8x(l)  =  (3)  Length  of  a  surface  of  72  sq.  ft.,  1  ft.  w.  =  72  ft. 

Note.— To  multiply  the  length,  (1)  multiply  the  area,  or  (2)  divide  the 
width. 

Since,  by  principle  2  above,  the  length  and  width  bear  to  the 
area  the  numerical  relation  of  factors  to  a  product,  the  numer- 
ical relation  of  the  area  and  one  dimension  to  the  other  dimen- 
sion may  be  expressed  as  follows : 

6.  The  area  and  one  dimension  hear  to  the  other  dimension  the 
same  numerical  relation  that  the  dividend  and  divisor  hear  to  the 
quotient. 

Question :  Does  Prin.  3,  Art.  27,  apply  to  example  12?    Explain. 

IS.  The  width  of  a  solid  of  600  cu.  ft.,  15  ft.  long,  5  ft.  thick 
=8  ft.     Multiply  by  5.     (3  forms.) 

Process  :  (1)  Width  of  a  solid  of  600  cu.  ft.,  15  ft.  1.,  5  ft.  th.  =8  ft. 
5x(l)  =  (2)  Width  of  a  solid  of  3000  cu.  ft.,  15  ft.  1.,  5  ft.  th.=40ft. ;  or, 
5x(l)  =  (3)  Width  of  a  solid  of  600  cu.  ft.,  3  ft.  l.,5ft.  th.=40ft. ;  or, 
5x(l)  =  (4)  Width  of  a  solid  of  600  cu.  ft.,  15  ft.  1.,  1  ft.  th.  =40  ft. 

Note.— To  multiply  one  dimension,  (1)  multiply  the  volume,  or  (2)  di- 
vide either  of  the  other  dimensions. 

7.  The  volume  and  two  dimensions  hear  to  the  third  dimension 
the  same  numerical  relation  that  the  dividend  and  the  factors  of  the 
divisor  bear  to  the  quotient. 

Question :  Does  Prin.  3,  Art.  27,  apply  to  example  13? 

14..  The  principal  required  to  gain  $36  in  6  years  at  6%  = 
$100.     Multiply  by  6.     (3  forms.) 


88  ADVANCED    ARITHMETIC. 

Process  :   (1)  Prin.  reqd.  to  gain  $36  in  6  yr.  at  6^  =  $100. 
6x(l)  =  (2)  Prin.  reqd.  to  gain  $216  in  6  yr.  at  6^  =  $600;  or, 
6x(l)  =  (3)  Prin.  reqd.  to  gain  $36  in  1  yr.  at  6^  =  $600;  or, 
6x(l)  =  (4)  Prin.  reqd.  to  gain  $36  in  6  yr.  at  1^  =  $600. 

Note.— To  multiply  the  principal,  (1)  multiply  the  interest,  or  (2)  di- 
vide the  time,  or  (3)  divide  the  rate. 

8.  The  interest  and  any  two  of  the  other  elements  (principal, 
rate,  or  time)  bear  to  the  third  element  the  same  numerical  relation 
that  the  dividend  and  the  factors  of  the  divisor  bear  to  the  quotient. 

Question:  Does  Prin.  3,  Art.  27,  apply  to  example  14? 

15.  ■  The  time  required  by  10  men  to  saw  10  cords  of  wood  =  1 
day.     Multiply  by  10.     (2  forms.) 

Process  :   (1)  Time  reqd.  by  10  m.  to  saw  10  c.  =  1  da. 
10x(l)  =  (2)  Time  reqd.  by  1  m.  to  saw  10  c.  =  10  da. ;  or, 
10x(l)  =  (3)  Time  reqd.  by  10  m.  to  saw  100  c.  =  10  da. 

Note. — To  multiply  the  time,  (1)  multiply  the  work  to  be  done,  or  (2) 
divide  the  working  force. 

9.  The  work  done  and  either  element  (working  force  or  time)  bear 
to  the  other  element  the  same  numerical  relation  that  the  dividend 
and  divisor  bear  to  the  quotient. 

EXERCISE  XLIII. 

1.  17-44-6  =  19.     Multiply  by  10. 

2.  1  bushel =4  pecks.     Multiply  by  80. 

3.  3  quarts =6  pints.     Multiply  by  12. 

4..  89.87  inches=l  meter.     Multiply  by  200. 

5.  1  ton  =  20  hundredweight.     Multiply  by  6. 

6.  $50-4x  A's  money  =  $30.     Multiply  by  18. 

7.  Cost  of  1  pound=8/.     Multiply  by  14.     (2  forms.) 

8.  Value  of  1  book=75/.     Multiply  by  11.     (2  forms.) 


ADVANCED   ARITHMETIC.  89 

9,  Wt. of  12 loads  =  150000 pounds.  MultiplybyS.  (2forms.) 

10.  Price  of  15  share8  =  $750.     Multiply  by  20.     (2  forms.) 

11.  8X9  =  72.     Multiply  by  7.     (2  forms.) 

12.  Area  of  a  surface  1  ft.  long,  1  ft.  wide=l  sq.  ft.  Multi- 
ply by  7.     (2  forms.)      ' 

13.  Volume  of  a  solid  1  ft.  long,  1  ft.  wide,  1  ft.  thick  =  1 
cu.  ft.     Multiply  by  18.     (3  forms.) 

U.  72^8=9.     Multiply  by  4.     (2  forms.) 

15.  Width  of  a  surface  of  30  sq.  ft.,  15  ft.  long=2  ft.  Mul- 
tiply by  15.     (2  forms.) 

16.  Length  of  a  surface  of  80  sq.  ft.,  2  ft.  wide  =  15  ft.  Mul- 
tiply by  5.     (2  forms.) 

17.  Time  required  for  $500  to  gain  $80  at  8%  =2  yr.  Multi- 
ply by  3.     (3  forms.) 

18.  Length  of  a  wall  12  ft.  high,  2  ft.  thick,  built  by  6  men 
in  8  days  =  600  ft.     Multiply  by  2.     (4  forms.) 

19.  Principal  required  to  produce  $80  at  8%  in  2  years  =  $500. 
Multiply  by  12.     ( 3  forms.) 

20.  Rate  required  for  $500  to  produce  $80  in  2  years=8%. 
Multiply  by  7.     (3  forms.) 

21.  Thickness  of  a  solid  of  30  cu.  ft.,  6  ft.  long,  5  ft.  wide= 
1  ft.     Multiply  by  10.     ( 3  forms.) 

62.  To  Divide  an  Equation. — The  Third  Applica- 
tion of  Division  is  employed  in  dividing  equations.  An  equa- 
tion is  divided  by  taking  the  same  part  of  each  of  its  members. 

General  Principles:  1.  Equals  divided  by  the  same  number 
give  equal  quotients. 

.2.  Dividing  each  term  of  a  member  of  an  equation  by  a  number 
divides  the  member  by  that  number. 

Note. — In  addition  to  these  general  principles,  all  but  one  of  the  spe- 
cial principles  given  in  Article  61  will  be  of  service  in  the  following  ex- 
amples and  exercise. 


90  ADVANCED   ARITHMETIC. 

-RXATVTPTjTSS. 

i.  21  bushels =84  pecks.     Divide  by  7. 
Note. — Remember,  that  to  divide  by  7  means  to  take  \  of  each  member. 

Process  :    (1)  21  bu.  =  84  pk. 
iof(l)  =  (2)  3bu.  =  12pk. 

2.  Cost  of  9  hats=$36.     Divide  by  9.     (2  forms.) 

Process :    (1)  Cost  of  9  hats  =  $36. 
i  of  (1)  =  (2)  i  of  cost  of  9  hats  =  $4 ;  or, 
i  of  (1)  =  (3)  Cost  of  1  hat  =  $4. 

Note. —  Observe  that  either  equation  (2)  or  (3)  is  the  correct  result, 
and  that  ^  of  cost  of  9  hats  =  cost  of  1  hat. 

5.25-15+80=40      Divide  by  5. 

Process:    (1)25-15+30  =  40. 
^of  (1)  =  {2)  5-3+6  =  8.     (Prin.  2.) 

Jf..  20Xmy  age  — 120  years =8  X  John's  age.     Divide  by  4. 

Process :    (1)  20  x  my  age  - 120  yr.  =  8  x  John's  age. 

i  of  (1)  =  (2)  5  X  my  age  -  30  yr.  =  2  x  John's  age. 

5.  12X18=216.     Divide  by  6.     (2  forms.) 

Process:    (1)  12x18  =  216. 
lof  (1)  =  (2)  -1^x18  =  2x18  =  36;  or, 
^of(l)  =  (3)  12xJ/  =  12x3  =  36.    (Prin.  2,  Art.  27.) 

6.  Area  of  a  surface  15  feet  long  and    12   feet  wide  =  180 
square  feet.     Divide  by  3.     (2  forms.) 

Process  :    (1)  Area  of  a  surface  15  ft.  1.,  12  ft.  w.  =  180  sq.  ft. 
\  of  (1)  =  (2)  Area  of  a  surface  5  ft.  1.,  12  ft.  w.  =  60  sq.  ft. ;  or, 
I  of  (1)  =  (3)  Area  of  a  surface  15  ft.  1.,  4  ft.  w.  =  60  sq.  ft. 

Note.— Show  how  Prin.  2,  p.  85,  and  Prin.  2,  Art.  27,  apply  to  exam- 
ple 6. 

7.  The  volume  of  a  solid  18  ft.  long,  12  ft.  wide,  and  6  ft. 
thick  =  1296  cu.  ft.     Divide  by  6.     (3  forms.) 


ADVANCED   ARITHMETIC.  91 

Process.:    (1)  Vol.  of  a  solid  18  ft.  L,  12  ft.  w.,  6  ft.  th.  =  1296  cu.  ft. 

I  of  (1)  =  (2)  Vol.  of  a  solid  3  ft.  1.,  12  ft.  w.,  6  ft.  th.  =216  cu.  ft. ;  or, 

^of  (1)  =  (3)  Vol.  of  a  solid  18  ft.  1.,  2  ft.  w.,  6  ft.  th.  =  216ca.  ft.;  or, 

i  of  (1)  =  (4)  Vol.  of  a  solid  18  ft.  1.,  12  ft.  w.,  1  ft.  th.  =216  cu.  ft. 
< 

Note.— Show  how  I*Hii,  3  (2),  Art.  61,  and  Prin.  2,  Art.  27,  apply  to  ex- 
ample 7. 

<9.  72-^12  =  6.     Divide  by  2.     (2 forms.) 

Process:    (1)  72-12  =  6. 
iof  (1)  =  (2)  ^2^-^12  =  36  +  12  =  3;  or, 
iof(l)  =  (3)  72-h(2x12)  =  72-h24  =  3.     (Prin.  4,  Art.  27.) 

9.  Rate  required  for  $500  to  produce  $750  in  15  years=10%. 
Divide  by  5.     (3  forms.) 

Process :    (1)  Rate  reqd.  for  $500  to  produce  $750  in  15  yr.  =  W. 
I  of  (1)  =  (2)  Rate  reqd.  for  $2500  to  produce  $750  in  15  yr.  =2^ ;  or, 
^  of  (1)  =  (3)  Rate  reqd.  for  $500  to  produce  $150  in  15  yr.  =  2% ;  or, 
I  of  (1)  =  (4)  Rate  reqd.  for  $500  to  produce  $750  in  75  yr.  =  2^. 

Note. —  Show  how  Prin.  8,  Art.  61,  and  Prin.  4,  Art.  27,  apply  to  ex- 
ample 9. 

10.  The  time  required  by  10  men  to  saw  120  cords  of  wood  = 
10  days.     Divide  by  5.     (2  forms.) 

Process :    (1)  Time  reqd.  for  10  m.  to  saw  120  c.  =  10  da. 
J  of  (1)  =  (2)  Time  reqd.  for  50  m.  to  saw  120  c.  =2  da. ;  or, 
J  of  (1)  =  (3)  Time  reqd.  for  10  m.  to  saw  24  c.  =  2  da. 

Note. — Show  how  Prin.  9,  Art.  61,  and  Prin.  4,  Art.  27,  apply  to  ex- 
ample 10. 

11.  Thickness  of  a  solid  of  480 <3U.  ft.,  10  ft.  long,  8  ft.  wide 
=6  ft.     Divide  by  2.     (3  forms.) 

Process  :    (1)  Thickness  of  a  solid  of  480  cu.  ft.,  10  ft.  1.,  8  ft.  w.  =6  ft. 
^  of  (1)  =  (2)  Thickness  of  a  solid  of  480  cu.  ft.,  10  ft.  1. ,  16  ft.  w.  i=  3  ft. ;  or, 
i  of  (,1)  =  (3)  Thickness  of  a  solid  of  240  cu.  ft.,  10ft.  1.,  8  ft.  w.  =  3  ft. ;  or, 
^  of  (1)  =  (4)  Thickness  of  a  solid  of  480  cu.  f t. ,  20  f 1. 1. ,  8  ft.  w.  =  3  ft. 

Question  :  Why  increase  the  width  in  (2)  ?  the  length  in  (4)  ? 

Note.— Show  how  Prin.  7,  Art.  61,  and  Prin.  4,  Art.  27,  apply  to  exam- 
ple 11. 


92  ADVANCED   ARITHMETIC. 

EXERCISE  XLIV. 

1.  18-12+24=80.     Divide  by  6. 

2.  420  hundredweight =21  tons.     Divide  by  7. 

3.  $560-40%  of  cost=$280.     Divide  by  40. 
4.18X30=540.     Divide  by  6.     (2  forms.) 
5.4X12X10=480.     Divide  by  2.     (Sforms.) 

6.  50%  of  75%  of  a  number  =  250.     Divide  by  25.     ( 2  forms.) 

7.  Weight  of  20  barrels =5000  pounds.     Divide  by  10.     (2 
forms.) 

8.  Cost  of  30  oranges =40/.     Divide  by  80.     (2  forms.) 

9.  The  volume  of  a  solid  12  ft.  long,  4  ft.  high,  8  ft.  thick 
=  884  cu.  ft.     Divide  by  4.     (3  forms.) 

10.  Interest  on  $6000  at  8%  for  IJ  yr.  =  $640.  Divide  by  8. 
(3  forms.) 

ii.  24-^8=3.     Divide  by  3.     (2forms.) 

12.  Time  required  by  14  women  to  make  7  dresses =3^  days. 
Divide  by  3^.     ( 2  forms.) 

13.  Width  of  a  surface  of  108  sq.  ft.,  12  ft.  long=9  ft.  Di- 
vide by  9.     (2  forms.) 

1j^.  Length  of  a  wall  12  ft.  high,  2  ft.  thick,  built  by  6  men 
in  12  days  =  900  ft.     Divide  by  6.     (4  forms.) 

15.  Principal  required  to  produce  $640  in  1^  years  at  8%  = 
$6000.     Divide  by  64.     ( 3  forms.) 

16.  Rate  required  for  $6000  to  produce  $640  in  1^  years =8% . 
Divide  by  4.     (3  forms.) 

17.  Time  required  for  $6000  to  produce  $640  at  8%  =  1^  years. 
Divide  by  4.     (3  forms.) 

2.     RATIO. 

63.  Definitions. — A  ratio  is  an  expression  of  the  rela- 
tive magnitude  of  one  number  as  compared  with  another  of  the 
same  kind  taken  as  the  standard.  The  number  compared  is 
the  antecedent ;  the  one  taken  as  the  standard  of  comparison  is 
the  consequent.  These  numbers  are  also  the  terms  of  the  ratio, 
and  together  form  a  couplet. 


ADVANCED   ARITHMETIC.  \)6 

15  yd.:  5  yd.  =3 

is  read,  **The  ratio  of  15  yd.  to  5  yd.  equals  3,"  and  means 
that  15  yd.  is  3  times  5  yd. 

'-     $8:  $12  =  1 

means  that  $8  is  |  of  $12.  Observe  that  a  ratio  is  always  ah- 
stract.  15  yd.  is  S  times  5  yd.,  but  not  3  yd.  times  5  yd. ;  $8  is 
f  of  $12,  but  not  $1  of  $12. 

In  arithmetic  we  have  the  indicated  ratio,  and  the  numerical 
ratio.     As, 

In     $32:  $16=2 

$32:  $16  (indicated  ratio). 
2  ( numerical  ratio). 

For  every  indicated  ratio  whose  terms  are  concrete  there  may 
be  written  a  corresponding  indicated  ratio  whose  terms  are  ab- 
stract.    As, 

70  bu.  :  14  bu.     (Terms  concrete.) 

70         :  14  (Terms  abstract.) 

Note. — The  respective  terms  in  the  two  ratios  are  numerically  equal ; 
the  ratios  are  absolutely  equal. 

Principle  :  The  numerical  ratio  between  any  two  similar  con- 
crete numbers  is  the  same  as  the  numerical  ratio  between  their  cor- 
responding abstract  numbers. 

64.  Writing  the  Corresponding  Abstract 
Terms  of  Ratios. — In  solving  ratios  and  proportions,  it 
is  best  to  deal  with  abstract  terms.  The  following  principles 
will  be  of  service  in  determining  the  corresponding  abstract 
terms  of  ratios  whose  terms  are  concrete. 

Note.— These  principles  are  found  in  Article  61,  but  are  not  grouped 
there  as  they  are  here. 


94  ADVANCED   ARITHMETIC. 

I.  Principles  Relating  to  Surfaces:  (1)  The  dimensions 
(length,  and  width)  hear  to  the  area  the  same  numerical  relation  as 
that  of  factors  to  the  product.  (2)  The  area  and  one  dimension 
hear  to  the  other  dimension  the  same  numerical  relation  that  the 
dividend  and  divisor  bear  to  the  quotient. 

II.  Principles  Relating  to  Solids:  (1)  The  dimensions 
(length,  width,  and  thickness)  hear  to  the  volume  the  same  numeri- 
cal relation  as  that  of  factors  to  the  product.  (2)  The  volume  and 
two  dimensions  bear  to  the  third  dimension  the  same  numerical  re- 
lation that  the  dividend  and  the  factors  of  the  divisor  bear  to  the 
quotient. 

III.  Principles  Relating  to  Interest:  (1)  The  principal, 
time,  and  rate  bear  to  the  interest  the  same  numerical  relation  as 
that  of  factors  to  the  product.  (2)  The  interest  and  any  two  of  the 
elements  (principal,  rate,  or  time)  bear  to  the  third  element  the 
same  numerical  relation  that  the  dividend  and  the  factors  of  the 
divisor  bear  to  the  quotient. 

IV.  Principles  Relating  to  Work:  (1)  The  working  force 
and  time  hear  to  the  work  done  the  same  numerical  relation  as  that 
of  factors  to  the  product.  (2)  The  work  done  and  either  of  the  ele- 
ments (working  force  or  time)  hear  to  the  other  element  the  same 
numerical  relation  that  the  dividend  and  divisor  hear  to  the  quotient. 

EXAMFIiES. 

The  following  are  examples  of  writing  corresponding  abstract 
terms  of  ratios : 

1.  70  bushels:  14  bushels. 

Abstract:  70:14. 

2.  5X20  men:  4x8  men. 

Abstract:  5x20:4x8. 

3.  \  of  cost  of  32  books :  f  of  cost  of  21  books. 

Abstract:  ^of  32:|of  21. 


8. 


ADVANCED    ARITHMETIC.  95 

^     Area  of  a  surface     }      {     area  of  a  surface     ) 
^-     \l2  ft.  long,  9  ft  wide  ^  *  ^  9  ft.  long,  8  ft.  wide  ]  ' 
Abstract ;  12  x  9 : 9  x  8.     (  Prin.  I-(l).) 

C   Length  of  a  solid  of   )      (    length  of  a  solid  of 
5.     \  72  cu.  ft.,  4  ft.  wide,  >  :  ^  90  cu.  ft.,  5  ft.  wide, 
(  2  ft.  thick  )      (  3  ft.  thick 

Abstract :  j^ :  g^^.    ( Prin.  II-(2).) 

{  Interest  on  $500  )      {  interest  on  $400  ) 
^'     I  at  6%  for  3i  yr.  ^  *  ^  at  8%  for  5  yr.  \' 
Abstract:  500x6x31:400x8x5.    (Prin.  III-(1).) 

C  Rate  required  for  )      C  rate  required  for  ) 
7.     <  $500  to  gain  $90    >  :  <  $400  to  gain  $160  >  . 
(  in  8  years  )      (  in  5  years  ) 

Abstract:  ^^::^^.    (Prin.  III-(2).) 


Work  done  by     }  ,   \      work  done  by 
15  men  in  12  days  ^  '   ^  11  men  in  8  days 

Abstract:  15x12:11x8.    (Prin.  IV-(1).) 


i     Number  of  men  re-     )      C     number  of  men  re- 
9.     <  quired  to  build  180  rd.  >  :  <  quired  to  build  88  rd. 
(  of  fence  in  12  days         )      (of  fence  in  8  days 

Abstract :  ^ :  f .    ( Prin.  IV-(2;.) 


EXEECISE  XLV. 

Write  the  corresponding  ratios  with  abstract  terms : 

1.  25  1b.:  15  1b. 

2.  8x9  books:  7X12  books. 

3.  Cost  of  10  shares :  cost  of  15  shares. 
4..   Value  of  20  loads :  value  of  14  loads. 

5.  7  X  my  age :  12  X my  age. 

6.  9 X cost  of  10  hats:  12 X cost  of  5  hats. 


96 


ADVANCED   ARITHMETIC, 


7. 
8. 
9. 

10. 
11, 
12. 

13. 

U. 
15. 
16. 
17. 

18. 

19. 


Area  of  a  surface 
10  rd.  long,  8  rd.  wide 

Length  of  a  surface  of 
160  sq.  ft.,  8  ft.  wide 

Width  of  a  surface  of 
380sq.  rd.,20rd.long 

Volume  of  a  solid 
20  ft.  long,  8  ft.  high, 
2  ft.  thick 

Length  of  a  solid  ) 
of  140  cu.  yd.,  7  yd.  \ 
high,  2  yd.  thick        ) 

Width  of  a  solid  ) 
of  120  cu.  yd.,  8  yd.  V 
long,  3  yd.  thick        ) 

Thickness  of  a  solid  ) 
of  108  cu.  ft.,  12  ft.  V 
long,  9  ft.  wide  ) 

Interest  on  $500  \  \ 
for  6  yr.  at  10%  ]  '   I 

Time  reqd.  for  $600 
to  gain  $75  at  6% 

Rate  reqd.  for  $540 
to  bear  $86.40  in  2  yr. 

Work  done  by 
12  men  in  25  days 

Time  reqd.  by  6  men 
to  build  40  rd.  of  fence 

Number  of  men  reqd.  ) 
to  build  140  rd.  in  18  [ 
days  ) 

Principal  reqd.  to      j 
produce  $112  in  2^  yr. 
at  7%  ] 


area  of  a  surface        ) 
17  rd.  long,  12  rd.  wide  ) 

)  length  of  a  surface  of  ) 
I  860  sq.ft.,  18  ft.  wide  \  ' 

width  of  a  surface  of  I 
540sq.rd.,30rd.  long  \  ' 

)  C  volume  of  a  solid  ) 
}  :  ^30  ft.  long,  10  ft.  high,  [ 
)      (  H  ft.  thick  ) 

C     length  of  a  solid     ) 

:  ]  of  120  cu.  yd.,  5  yd.  S  . 

(  high,  3  yd.  thick        ) 

(      width  of  a  solid      ) 

:  ^of  140cu.  ft.,  10  ft.  }  . 

(  long,  2  ft.  thick         ) 

(  thickness  of  a  solid  ) 

:  ]  of  120  cu.  ft.,  10  ft.  [  . 

(  long,  8  ft.  wide  ) 

interest  on  $750 
for  3  yr.  at  8% 

<  time  reqd.  for  $480  ) 
'   \  to  gain  $60  at  10%   ) 


\    rate  reqd.  for  $620   ) 
'    I  to  bear  $248  in  4  yr.  ) 


work  done  by 
16  men  in  20  days 

time  reqd.  by  15  men     ) 
to  build  120  rd.  of  fence  ) 

number  of  men  reqd.  ) 
to  build  320  rd.  in  15  L 
days  ) 

principal  reqd.  to      ) 
produce  $120  in  3^  yr.  y  . 
at  5%  ) 


ADVANCED    ARITHMETIC.  97 

65.  The  Process  of  rinding"  tlie  Numerical 
Ratio. — The  process  employed  in  determining  the  numerical 
ratio  between  two  numbers  is  division.  (  See  Second  Applica- 
tion, Art.  16.)  The  antecedent  becomes  the  dividend;  the 
consequent,  the  divisor  |  and  the  numerical  ratio,  the  quotient. 

Principles:  1,  The  antecedent  divided  by  the  consequent  eqiiah 
the  numerical  ratio, 

2.  When  the  terms  are  abstract^  the  antecedent  divided  by  the  nu- 
merical ratio  equals  the  consequent. 

3.  The  product  of  the  consequent  multiplied  by  the  numerical  ra- 
tio equals  the  antecedent. 

EXAMPLES. 

1.  The  antecedent  is  $20,  the  consequent  $5.  Find  the  nu- 
merical ratio. 

Statement:  $20:$5  =  (  )? 
Process:  20-^5  =  4. 

Note. — It  is  recommended  that,  in  the  process ,  the  corresponding  ab- 
stract terms  be  always  used ;  then,  no  trouble  will  arise  in  dealing  with 
complicated  expressions,  such  as  are  found  below  (see  "No.  7). 

The  statement  is  made  for  the  purpose  of  showing  clearly  the  relation 
of  the  given  parts  to  the  required  part, 

2.  12  is  the  ratio  of  72  days  io  what  ? 

Statement :  72  da. :  (  )  =  12  ? 
Process:  72+12  =  6.    (Prin.  2.) 

.'.  6  days  is  the  consequent. 

Question :  Why  is  the  consequent  days  ? 

3.  10  is  the  ratio  of  what  number  to  13  men  ? 

Statement :  (  ) :  13  men  =  10. 

Process:  10x13  =  130.    (Prin.  3.) 

.*.  130  men  is  the  antecedent. 


98  ADVANCED    ARITHMETIC. 

Jf.  Find  the  ratio  of  the  work  of  6  men  for  15  days  to  the 

work  of  9  men  for  2  days. 

Qt^in^.,^4  .   S  Work  of  6  men  \  .  S  work  of  9  men  ?  _/  x  o 
statement :   j      ^^^,  ^^  ^^^^     ^  :  j      ^^^  ^  ^^^^     |  =n  ? 

Abstract:  6xl5:9x2  =  (  )? 
6x15    , 


Process 


9x2 


5.  One  block  of  marble  is  10  ft.  long,  5  ft.  wide,  2  ft.  thick ; 
another  is  15  ft.  long,  4  ft.  wide,  5  ft.  thick.  The  volume  of 
the  first  bears  what  ratio  to  the  volume  of  the  second  ? 


volume       ^ 
statement:  (1)  <   Ift.'w!,  \  ■  j   Vu.w.,  [  =<  )? 


Process.    ^-     .     _ 
15  X  4  X  5 


i 


5  ft.  th. 
15  X 
4x    H(  )? 


Note. — Employ  cancellation  wherever  convenient. 

6.  A  tract  of  land  50  rd.  long,  30  rd.  wide,  is  1^  times  as  large 

as  another  tract,  45  rd.  long.     How  wide  is  the  second  tract  ? 

(     Tract       )       (       tract       ) 
Statement:  (1)  <    50  rd.  1.,    }  :  ]    45  rd.  1.,    ^=1}? 
(    30  rd.  w.  )       (    (  )  rd.  w.   ) 

(2)||0X    \..\IY    \-m 

50x30    __ 
Process:  -rz — 77  =  25. 

.'.  the  required  width  is  25  rd. 

7.  What  is  the  ratio  of  the  principal  required  to  produce 
$216  interest  in  5  yr.  at  6%  to  the  principal  required  to  pro- 


duce $192  in  4  yr.  at 


iPrin.  reqd.  to    )      (    prin.  reqd.  to    ) 
produce  $216  in  V  :  -^  produce  $192  in  [•  =(  )? 
5  yr.  at  6^  )      (  4  yr.  at  8^  ) 


216^.^92^ 
^  ''5x6*4x8    ^  ^' 


ADVANCED    ARITHMETIC. 


99 


216x4x8     .     ., 
^''^''''•'   5x6x192  =  ^==^^- 


8,  Find  the  ratio  of  the  length  of  a  surface  of  144  sq.  ft.,  12 


ft.  wide,  to  the  length  "pf  a  surface  of  180  sq.  ft.,  10  ft.  wide 

!  Length  of  a  surface  )  (  length  of  a  surface  ) 
of  144  sq.  ft.,  12  ft.  \  :  ]  of  180  sq.  ft.,  10  ft.  \ 
wide  )      ( wide  \ 


Statement:   (1) 


(  )? 


(2) 


144,180^ 
12  *  10     ^ 


Process 


144x10 
12x180 


=  |. 


EXERCISE  XLVI. 


Find  the  numerical  ratios  in  the  following 


FIND  THE  BATIO. 

FIND  THE  CONSEQUENT. 

FIND  THK  ANTECEDENT. 

Antecedent. 

Consequent. 

Antecedent. 

Ratio. 

Consequent. 

Ratio. 

1.  56  ft. 

8  ft. 

7.  $64. 

11. 

13.  342  in. 

1 
a* 

2.  540  yd. 

18  yd. 

8.  16  da. 

h 

U.  34  mi. 

f. 

3.  2760  men. 

60  men. 

9.  98  sq.  ft. 

10. 

15.  72  yr. 

13. 

4-  H- 

$20. 

10.  $7000. 

140. 

16.  .003  oz. 

440. 

5.  34  bu. 

16  bu.      1  11.  i  hr. 

72. 

17.  64  cords. 

9. 

6.  12  hr. 

54  hr.       !  12.  24  gal. 

u- 

18.  til  meters. 

1852. 

Find  the  numerical  ratios  in  Exercise  XL  V. 

8.     PROPORTION. 

66.  Definitions. — A  proportion  is  the  expression  of 
equality  between  two  indicated  ratios.  Every  proportion, 
therefore,  has  4  terms,  which  are  numbered  in  order  from 
left  to  right.  The  first  and  fourth  terms  are  the  extrenus;  the 
second  and  third  terms,  the  means.  The  double  colon  (::  )  is 
usually  (but  not  always)  used  for  the  sign  of  equality  between 
the  ratios. 

8bu. :  3bu. ::  $40:  $15, 

is  read,  **8  bu.  is  to  3  bu.  as  $40  is  to  $15,"  and  means  that 
the  ratio  of  8  bu.  to  3  bu.  is  equal  to  the  ratio  of  $40  to  $15. 


100  ADVANCED    ARITHMETIC. 

67.  Denominations  of  Terms. — (1)  All  the  terms 
of  a  proportion  may  be  concrete.     As, 

$25 :  $12 : :  100  men  :  48  men. 

Remember,  that  the  two  terms  of  an  indicated  ratio  must  be  similar ; 
but  terms  in  different  ratios  need  not  be  similar,  even  though  those  ra- 
tios are  in  the  same  proportion. 

(2)  One  couplet  may  be  concrete  and  the  other  abstract.    As, 

■  $25:  $12::  100: 48. 

(3)  All  the  terms  may  be  abstract.     As, 

25:  12::  100:48. 

By  dropping  the  denominations  in  a  proportion,  all  or  a  part 
of  whose  terms  are  concrete,  a  new  proportion  is  formed  whose 
terms  are  abstract  but  are  numerically  equal  to  those  of  the 
first  proportion.     Thus, 

(1)  80  yr. :  12  yr. : :  90  bu. :  86  bu.     (Terms  concrete.) 

(2)  80 :  12 : :  90 :  86.     (Terms  abstract.) 

68.  Solving"  a  Proj^ortion  with  Abstract 
Terms. —  Solving  a  proportion  is  the  process  of  finding  any 
one  of  its  terms  when  the  other  three  are  known. 

Principles:  1.  In  any  'proportion  whose  terms  are  abstract^  the 
product  of  the  extremes  is  equal  to  the  product  oj  the  means. 

Thus,  (1)  7: 15::  21:  45. 
Then,  7x45  =  15x21. 

(2)  8:3::  32: 12. 
Then,  8x12  =  3x32. 

2.  Either  extreme  is  equal  to  the  product  of  the  means  divided  by 

the  other  extreme. 

Thus,  20:  34::  30:  51. 

„,         34x30    _,         ,34x30    ^ 
Then,      ^     =51,  and     ^^     =20. 


ADVANCED    ARITHMETIC.  101 

3.  Either  mean  is  equal  to  the  product  of  the  extremes  divided  by 
the  other  mean. 

Thus,  36: 14::  18:  7. 

.  mu        36x7    ,o       J  36x7    , . 
Then,     ..  ^    =18,  and    ..  „    =14. 
14  18 


EXAlCFIiES. 


1.  Find  the  missing  term :  12 :  25 : :  (   )  :  125  ? 

„  12x125    _^ 

Process  :  — — —  =  60. 

Note. — Use  cancellation  where  convenient. 


?.  Find  the  missing  term :   (   ) :  82 : :  60 :  8  ? 


„  32x60    „.„ 

Process  :  —x —  =  240. 
8 

3.  Find  the  missing  term :   13 :  12 : :  78 :  (  )  ? 

12x78    -„ 
Process  :  — i-r —  =  72. 

4.  Find  the  missing  term :  9 :  (   )  : :  72 :  4  ? 

D  9x4    , 

Process  :  -fj^^h 

5.  Find  the  missing  factor:  (12x6):  35: :(  )X8:70? 

T    „  ...  12x6x70    ... 

I.  Process:  (1)  — =  144. 

do 

(2)  144  +  8  =  18. 

^  TT    r,  12x6x70    ,„ 

Or,  II.  Process :      g^^g     =18. 

Note.— The  plan  of  the  first  process  is  to  neglect  the  8  and  solve  for 
the  entire  third  term  (144) ;  then,  dividing  by  8  will  give  the  required 
factor.  The  plan  of  the  second  process  is  this :  Since  the  product  of  all 
the  factors  of  the  extremes  equals  the  product  of  all  the  factprs  of  the 
means,  the  product  of  all  the  factors  of  the  extremes  divided  by  the 
product  of  all  but  one  of  the  factors  of  the  means,  will  give  that  one. 


102 


ADVANCED    ARITHMETIC. 


^    120X100    108x100      ,^    ^„ 
^-     (    )X6   ''    720X5   ••12:9? 

p  .       108x100x12x6 J_ 

^^^^^^*   720x5x120x100x9  "500" 
.*.  the  missing  factor  is  500. 

Note. — Where  the  fractions  are  to  be  multiplied  together,  their  nu- 
merators are  placed  above  and  their  denominators  are  placed  below  the 
line  (why  ?  ) ;  when  used  as  divisors,  their  denominators  are  placed  above 
and  their  numerators  below  the  line  (why  ?  ).  Special  attention  should 
be  given  to  the  fact,  that  if  the  required  factor  is  in  a  denominator,  it 
will  appear  in  the  result  as  a  denominator  with  1  for  its  numerator. 


7. 


128 
16 


12_ 
(   ) 


::25:45? 


Process  : 


128x45 


=  h 


16x72x25 
,*.  the  required  factor  is  5  (in  the  denominator) 


1.  12:3 

2.  12:3::4 

3.  (   ):72 

4.  43:  (   ) 

5.  221: 13 

6.  {):li 

7.  A:(   ) 


(   ):4? 

(   )? 
li:3? 
20:600? 
(   ):21? 
360:720? 
4:9? 


EXERCISE  XLVII. 
125 


8.  9|:4::360:  (  ), 


9. 


10. 


12. 


46X( 
8X5 


) 


86 


25X24 


11.  27 


19x7'9x(   )•• 
12X6       108 
19 


14x28" 
162  210 


540X5*  700x6* 


•  175* 
12 :  72  ? 

•  49  * 
:(    ):5? 


69.  Solving  a  Proi^ortion  with  Concrete 
Terms. — In  any  proportion  whose  terms  are  concrete,  a  miss- 
ing part  may  be  found  as  follows : 

First,  form  and  solve  the  corresponding  proportion  whose  terms 
are  abstract ;  then,  determine  from  the  given  proportion  by  inspec- 
tion the  denomination  of  the  required  term. 


ADVANCED    ARITHMETIC.  103 

EXAICFXiES. 

Find  the  missing  part : 

1.  7  da. :  12  da. : :  (   )  mo. :  72  mo„? 

A  bstract  tef ms  ;  7 :  12 : :  (  ) :  72  ? 

„    '     '7x72     ._ 
Process  >  =  42. 

.*.  42  mo.  is  the  required  term. 

(  Area  of  a  surface  )      (  area  of  a  surface  ) 

2.  \  40  rd.  long,  20     [  :  ]  80  rd.  long,  (    )  V  : :  5  A. :  30  A.? 
/  rd.  wide  \      I  rd.  wide  ) 


Abstract  terms  :    20x40:  80x(  )::  5:  30  ? 
20x40x30     „ 
80x5      =^^- 
60  rd.  is  the  required  part. 


„  20x40x30    „ 

Process:  -^^^^  =  m. 


iSOO  reams  of 
16  pages  to 
the  sheet 


)      (  (  )  reams  of  )        (  38400  vol.  )      (  27000  vol.  ) 
[:  ]24:  pages  to  V  ::  ^  of  160  pp.  ^  M  of  320  pp.  [  ? 
)      (    the  sheet    )       (      each      )      (      each      ) 

800x16x27000x320    „„ 

^<"^^'«-        24X38400X160 ^^- 

.*.  750  reams  is  the  required  term. 

iBody  of  soldiers  )      (  body  of  soldiers  ) 
(    )  men  long,  60  V  :  -^  100  men  long,  75  [•  : :  No.  of  men :  itself  ? 
men  wide        )      (       men  wide       ) 

Abstract  terms  :  (  )  x  60 :  100  x  75 : :  No. !  No.  ? 
Process-  lQQx75xNo.^ 
60  X  No. 
o*.  125  men  is  the  required  part. 

Note. — The  **  No,"  being  in  both  numerator  and  denominator,  cancels. 

STime  reqd.  by     )      (    time  reqd.  by    ) 
(   )  men  to  build  >  :  -^  8  men  to  build  \  : :  34:  170  ? 
102  rd.  of  fence    )      {  680  rd.  of  fence  ) 

Abstract  terms  ;  ^ :  ^: :  34 :  170  ? 

V  )        8 


104  ADVANCED    ARITHMETIC. 

„  680x34 

.'.6  men  is  the  required  part. 

Note. — Do  not  forget  to  use  the  reciprocal  of  the  result,  when  the  re- 
quired term  is  a  factor  of  the  denominator. 

C    Principal  reqd.  to    )       (  principal  reqd.  to  ) 
6.  \  gain  $112  in  (   )  yr.  I  :  -^  gain  $120ineSiyr.  V  ::  8:  9  ? 
/at  7%  \      I  at  5%  j 

112       120 
Abstract  terms  :  r-^ — ^  •  tt. — ^ : :  8 :  9  ? 
(  )x7   3^x5 

„  120x8x7x3      ^ 

^^^^^^^•'112X5X9X10=^- 

.*.  I  or  2|  yr.  is  the  reqd.  part. 

EXERCISE  XLVIII. 
Find  the  missing  part  in  each  of  the  following  : 

1.  $120:  (  ) : :  48  cords :  20  cords  ? 
Note.— The  2d  term  must  be  $'s ;  why  ? 

2.  $3.50:  $50::  8  yd. :(   )  ? 

Note. — The  4th  term  must  be  yards ;  why  ? 

3.  44  in.:  18  in. : :  (  ) :  900  in.  ? 
of  a  solid      )      (     vol.  of  a  solid     'J 


(      Vol. 

:.   I    20  ft. 

(wide,  I 


long,  8  ft.    >  :  <  80  ft.  long,  10  ft.  V  : :  82 :  45  ? 
(   )  ft.  thick  \      I  wide,  1^  ft.  thick) 

(  Length  of  a  solid  ')      C  length  of  a  solid  ^ 

5.  \  140  cu.  yd.,  7  yd.  f  :  -^  (    )  cu.  yd.,  5  yd.  ^  : :  20:  16  ? 
(high,  2  yd.  thick  J      (high,  8  yd.  thick   ) 

f  Width  of  a  solid  ")      (    width  of  a  solid   ^ 

6.  \   150  cu.  ft.,  9  ft.    \  :  \  180  cu.  ft.,  (  )  ft.  f  : :  25: 18  ? 
(  high,  2  ft.  thick  )      (  high,  8  ft.  thick    ) 

(  Thickness  of  a  solid)      (  thickness  of  a  solid") 

7.  \    162  cu.  ft.,  12  ft.     [  '.  \     216  cu.  ft.,  Oft.     f::():2? 
(     long,  9  ft.  high      )      (    long,  8  ft.  high      ) 


ADVANCED   ARITHMETIC.  105 


5:8? 


j  Interest  on  $500  )      j    interest  on  $720   )     . 
^'  I  for  6  yr.  at  10%  J  '  (  f or  (   )  yr.  at  8%  J  "  * 

(  Time  reqd.  by  6  )      (  time  reqd.  by  15  i 
9.  \  men  to  biiild;40  [  :  I  men  to  build  120  V  : :  5 :  (   )  ? 
(  rd.  of  fence        ')      {  rd.  of  fence  ) 

{Number  of  men  reqd.  )      (  number  of  men  reqd.  ) 
to  build  320  rd.  of     [■  :  ^     to  build  140  rd.  of     [•  : ;  96 :  35  ? 
fence  in  15  days       j      (      fence  in  (  )  days       ) 


B.     PROBLEMS   OF   ONE   BASIS —NATURE   AND 
CLASSIFICATION. 

70.  Nature. — Typically,  a  problem  consists  of  two  parts: 
(1)  A  question,  proposed  for  solution;  and  (2)  a  statement  of 
a  certain  condition  or  relation,  from  which  that  solution  may  be 
determined.  These  parts  are  called  Question  and  Basis. 
Thus, 

If  6  books  cost  $2.40,  what  cost  10  books  ? 

Parts:  (1)  What  cost  10  books  ?     (Question.) 
(2)  6  books  cost  $2.40.     (Basis.) 

Note. — The  above  description  refers  only  to  such  problems  as  require 
but  one  unknown  part  or  term  to  be  found.  One  basis  only  is  needed  for 
the  solution  of  such  a  problem.  In  Part  III,  particular  attention  will 
be  given  to  problems  of  two  or  more  bases. 

Comparatively  few  problems  are  stated  in  the  form  of  the 
type ;  yet,  all  problems  have  these  two  essential  parts,  expressed 
or  implied. 

The  following  are  some  of  the  forms  not  typical : 

1.  Find  I  of  200  yards. 

2.  150  ounces  is  f  of  what  ? 
S.  Reduce  40000  lb.  to  tons. 
^.  $210  is  what  part  of  $450  ? 

5.  Find  the  area  of  a  floor  20  ft.  long  and  15  ft.  wide. 

6.  What  cost  6  doz.  eggs  at  20/  per  doz.? 


106  ADVANCED    ARITHMETIC. 

These  may  all  be  reduced  to  the  form  of  the  type.     Thus, 

1.  Since  all  {|)  of  200  yd.  is  200  yd.,  what  is  |  of  it  ? 

2.  If  I  of  a  number  is  150  ounces,  what  is  the  number  ? 

3.  Since  2000  lb.  make  1  ton,  how  many  tons  in  40000  lb.? 

4.  Since  $450  is  all  of  $450,  $210  is  what  part  of  it  ? 

5.  Since  the  area  of  a  surface  1  ft.  long,  1  ft.  wide  is  1  sq. 
ft.,  what  is  the  area  of  a  floor  20  ft.  long  and  15  ft.  wide  ? 

6.  If  1  doz.  eggs  cost  20/,  what  cost  6  doz.? 

71.  Stating  the  Parts  of  a  Problem. — Each  part 
of  a  problem  may  be  stated  in  the  form  of  an  eq^uation. 
Take  for  example  the  parts — 

(1)  What  cost  10  books  ?     (Question.) 

(2)  6  books  cost  $2.40.     (Basis.) 

The  question  may  be  stated  as  an  equation  between  cost  of 
10  books  and  the  required  term.     Thus, 

Cost  of  10books=(   )  ? 

In  like  manner,  the  basis  may  be  stated  as  an  equation  be- 
tween cost  of  6  hooks  and  $2.J^0.     Thus, 

(1)  Cost  of  6  books  =  $2.40;  or, 

(2)  $2.40=cost  of  6  books. 

Problems  often  involve  the  relation  of  more  than  two  num- 
bers. The  basis  may  have  as  many  forms  as  it  has  different 
numbers  related.     The  following  is  an  example : 

If  15  men  harvest  200  acres  of  grain  in  10  days,  in  how  many 
days  can  5  men  harvest  80  acres  ? 

The  relation  here  is  among  men  (working  force),  time,  and 
work  done  ;  and  its  three  forms  are : 

(1)  Amt.  harvested  by  15  men  in  10  days =200  A.;  or, 

(2)  Time  reqd.  for  15  men  to  harvest  200  A.  =  10  days;  or, 

(3)  Force  reqd.  to  harvest  200  A.  in  10  days  =  15  men. 


ADVANCED    ARITHMETIC.  107 

The  question  is  stated  as  before : 

Time  reqd.  for  5  men  to  harvest  80  A.  =  (   )  da.? 

The  following  principles,  which   have  been  illustrated  and 
explained,  should  be  learned : 

Principles:   /.  Every  problem  has  at  least  two  parts,  a  question 
{or  requirement)  and  a  basis. 

2.  The  parts  of  a  problem  may  always  be  stated  in  the  form  of 
equations. 

3.  Every  basis  has  as  many  forms  as  it  has  numbers  related. 

In  each  of  the  following  examples,  the  question  and  all  forms 
of  the  basis  are  stated  in  equations. 

EXAMPIjES. 

1.  James  has  $720,  and  John  has  |  as  much.     How  much  has 

John  ? 

Parts  :    (1)  |  of  James's  money  =  $(  )  ?    (  Question.) 
(2)  I  of  James's  money  =  $720  j  or, 

$720  =  I  of  James's  money.    ( Basis.) 

2.  What  will  10  hats  cost,  at  $2.40  each  ? 

Parts  :    (1)  Cost  of  10  hats  =  $(  )  ?    ( Question.) 
(2)  Cost  of  1  hat  =  $2.40;  or, 
$2.40  =  costof  Ihat.    (Basis.) 

3.  15%  of  a  debt  is  $630.     Find  the  debt. 

Parts:    (1)  100^  of  debt  =  $(  )?    (Question.) 
(2)  15^  of  debt  =  $630;  or, 

$630=  15^  of  debt.    (Basis.) 
Note. — The  sign, ^,  is  re&d  percent  and  means  hundredth  or  hundredths. 
10^  =  ^1^0^ ,  25^  =  i%% ,  etc.    100%  of  anything  is  all  of  it. 

4..  In  ^  of  a  load  of  corn,  there  are  6  bushels.     How  many 
bushels  in  25  such  loads  ? 

Parts:    (1)  Amount  of  25  loads  =  (  )  bu.  ?    (Question.) 
(2)  Amount  of  i\  of  a  load  =  6  bu.  ;  or, 

6  bu.  =  amount  of  i\  of  a  load.    ( Basis.) 


108  ADVANCED   ARITHMETIC. 

5.  f  of  an  article  is  worth  $.75.     What  is  W  of  it  worth  ? 

Parts  :    (1)  Value  of  \  I  of  article  =  $(  )  ?    ( Question.) 
(2)  Value  of  |  of  article  =  $.75;  or, 
$.75  =  value  of  |  of  article.    (Basis.) 

6.  If  6  men  plow  45  acres  in  3  days,  how  many  acres  will  1 
man  plow  in  1  day  ? 

Parts:    (1)  Amt.  plowed  by  1  man  in  Ida.  =  (  )  A.?    (Question.) 
(2)  Amt.  plowed  by  6  men  in  3  da.  =45  A. ;  or, 
Force  reqd.  to  plow  45  A.  in  3  da.  =  6  men ;  or, 
Time  reqd.  for  6  men  to  plow  45  A.  =  3  da.    ( Basis.) 

7.  If  25  boxes  of  pens  are  worth  $7.50,  what  is  ^  of  a  box 
worth  ? 

Parts  :    (1)  Value  of  J  box  =  $(  )  ?    (Question.) 
(2)  Value  of  25  boxes  =  $7.50;  or, 
$7.50  =  value  of  25  boxes.    (Basis.) 

8.  The  volume  of  a  solid  7  ft.  long,  5  ft.  wide  and  4  ft. 
thick  is  140  cubic  feet.  Find  the  volume  of  a  solid  10  ft. 
long,  7  ft.  wide  and  5  ft.  thick. 

Parts  :    (1)  Vol.  of  a  solid  10  ft.  1.,  7  ft.  w.,  5  ft.  th.  =  (  )cu.  ft.?  (Question.) 
(2)  Vol.  of  a  solid  7  ft.  1.,  5  ft.  w.,  4  ft.  th.  =  140  cu.  ft. ;  or, 
Length  of  a  solid  of  140  cu.  ft. ,  5  f t.  w. ,  4  ft.  th.  =  7  ft. ;  or. 
Width  of  a  solid  of  140  cu.  ft.,  7  ft.  1.,  4  ft.  th.  =  5  ft. ;  or, 
Thicknessof  asolidof  140cu.ft.,7ft.l.,5ft.w.  =  4ft.    (Basis.) 

9.  21  is  what  part  of  56  ? 

Parts:    (1)  21  =  (  )  of  56  ?    (Question.) 
(2)  56  =  all  of  56 ;  or. 

All  of  56  =  56.     (Basis.) 

Note.—**  21  =  (  )  of  56  ?  "  should  be  read,  **  21  equals  what  part  of  56  ?  " 

10.  760  is  how  many  times  38  ? 

Parts:    (1)  760  =  (  )x38  ?    (Question.) 
(2)  38  =  1x38;  or, 

1x38  =  38.     (Basis.) 


ADVANCED    ARITHMETIC.  109 

EXERCISE  XLIX. 

State  the  question  and  all  forms  of  the  basis  in  each  of  the  fol- 
lowing : 

1.  Find  20%  of  $600: 

2.  What  cost  f  of  a  ton  of  coal  at  $3.80  per  ton  ? 

3.  Clara  is  12  years  old ;  her  mother  is  four  times  as  old. 
How  old  is  her  mother  ? 

4.  Find  800%  of  $240. 

5.  I  of  my  salary  is  $875.     What  is  my  salary  ? 

6.  25%  of  a  certain  debt  is  $1600.     Find  the  debt. 

7.  If  14^  of  a  boat  is  worth  $9900,  what  is  the  whole  boat 
worth  ? 

8.  8i%  of  a  certain  sum  is  $50.  What  would  400%  of  that 
sum  be  ? 

9.  If  I  of  a  bolt  of  cloth  is  worth  $8.20,  what  are  18  such 
bolts  worth  ? 

10.  If  J  of  Mr.  Brown's  capital  is  $700,  how  much  is  f J  of  it  ? 

11.  I  deposit  48%  of  my  money,  which  is  $960,  in  the  bank, 
and  invest  28%  of  my  money  in  calves.  How  much  do  I  spend 
for  calves  ? 

12.  860%  of  a  number  is  252.     Find  the  number. 

13.  If  20  bu.  of  apples  cost  $18,  find  the  price  per  bushel. 
14..  If  5  men  can  saw  20  cords  of  wood  in  4  days,  how  many 

cords  can  1  man  saw  in  1  day  ? 

15.  120%  of  a  number  is  72.     Find  85%  of  it. 

16.  If  11  men  can  set  660  pages  of  type  in  6  days,  how  many 
pages  will  21  men  set  in  8  days  ? 

17.  240%  of  my  age  is  84  years.     Find  880%  of  it.  • 

18.  75  bu.  of  wheat  are  worth  $60.     What  are  42  bu.  worth  ? 

19.  A  lawyer  charges  $85  for  collecting  $700.  What  %  does 
he  charge  ? 

20.  I  bought  an  article  for  $5  and  sold  it  so  as  to  lose  $8. 
What  part  of  my  investment  did  I  lose  ? 


110  ADVANCED    ARITHMETIC. 

21.  I  bought  an  article  for  $5  and  sold  it  for  $7.50.  The 
selling  price  is  what  %  of  the  cost  price  ? 

22.  5  times  my  money  would  be  $4620.     How  much  have  I  ? 

23.  50  doz.  eggs  sell  for  $4.50.     Find  the  price  per  dozen. 
2J^.  Eggs  are  9/  per  dozen.     How  many  dozen  can  I  buy  for 

$8.60  ? 

25.  Eggs  are  9/  per  dozen.     What  will  80  dozen  cost  ? 

26.  Reduce  64  pints  to  quarts.     (Basis  omitted.) 

27.  The  area  of  a  surface  1  ft.  long,  1  ft.  wdde  is  1  sq.  ft. 
Find  the  length  of  a  lot  150  ft.  wide,  containing  30000  sq.  ft. 

28.  If  5  boys  plow  12  acres  in  IJ  days,  in  how  many  days 
will  12  boys  plow  60  acres  ? 

29.  Find  the  volume  of  a  stone  12  ft.  long,  6  ft.  wide,  and  8 
ft.  thick.     (Basis  omitted.) 

Note. — Preserve  your  work  for  use  in  Exercises  L  and  LI. 

72.  Classification. — The  problems  of  one  basis,  though 
varied  in  their  forms  of  expression,  when  considered  with  ref- 
erence to  the  relation  of  the  given  'part  to  the  required  part, 
consist  of  no  more  than  ten  different  types  ;  and  these  may  be 
classed  into  four  classes,  as  follows : 

Class  1.  Given  a  number,  to  find  (1)  a  part  of  it,  or  (2)  a  mul- 
tiple of  it. 

EXAMPLES. 

1.  What  is  the  value  of  f  of  an  article  which  cost  $7.20  ? 

2.  What  will  50  books  cost,  at  $2.50  each  ? 

Class  2.  Given  a  part  of  a  number,  to  find  (1)  the  mimber, 
(2)  a  multiple  of  it,  or  (8)  apart  of  it. 

EXAlttPIiES. 

1.  20%  of  a  sale  is  $540.  What  was  the  whole  amount  of 
the  sale  ? 

Note. — Any  number  of  %  less  than  100^  is  a  part ;  any  number  of  ^ 
greater  than  100^  is  a  multiple. 


ADVANCED   ARITHMETIC.  Ill 

2.  ^^  of  a  certain  farm  is  25  acres.  How  many  acres  in  8 
such  farms  ? 

S.  I  of  a  certain  article  is  worth  $.75.     What  is  f  of  it  worth  ? 

Class  3.  Given  h'^iultiple  of  a  number,  to  find  (1)  the  number , 
(2)  a  part  of  the  number,  or  (3)  another  multiple  of  it. 

EXA]SO>I.ES. 

1.  If  35  hats  sell  for  $105,  find  the  price  of  1  hat. 

2.  At  the  rate  of  $120  in  2  months,  how  much  money  can  be 
earned  in  f  of  a  month  ? 

S.  12  pounds  troy  equal  144  ounces ;  how  many  ounces  do  7 
pounds  troy  equal  ? 

Class  4.  Given  two  numbers,  to  find  (1)  what  part  one  is  of  the 
other,  or  (2)  what  multiple  one  is  of  the  other. 

EXAMPIiES. 

1.  A  commission  merchant  charges  $96  for  making  a  sale  of 
$3200.     What  per  cent  does  he  charge  ? 

2.  John  has  $.50  and  George  has  $5.  George's  money  is  how 
many  times  John's  ? 

It  may  be  ascertained  to  which  of  the  above  10  types  a  prob- 
lem belongs  by  examining  its  parts.  Thus,  in  example  2, 
Class  3, 

Parts:  (1)  Earnings  for  f  mo.  =  $(   )?     (Qaestion.) 
(2)  Earnings  for  2  mo.  =$120.     (Basis,) 
Here  are  given  the  earnings  for  2  months,  or  2  times  a  num- 
ber; to  find  the  earnings  for  f  month,  or  f  of  the  number.     It 
gives  a  multiple  of  a  number,  to  find  a  part  of  it. 

EXERCISE  L. 

1.  Determine  to  which  of  the  10  types  each  of  the  examples 
under  Arti<5le  71  belongs. 

2.  Determine  to  which  of  the  10  types  each  problem  of  Ex- 
ercise XLIX  belongs. 


112  ADVANCED    ARITHMETIC. 

C.     PROBLEMS  OF  ONE  BASIS  — SOLUTION. 

73.  Two  Metliods  of  Solution.— Solving  a 
Problem  consists  in  commencing  with  a  given  (or  known) 
relation  and  passing  by  means  of  an  authorized  process  to  the  re- 
quired relation.  There  are  two  methods  employed  in  arithme- 
tic for  solving  problems  of  one  basis : 

1.  The  Equation  Metliod,  in  which  equations  are  used 
throughout  to  express  the  steps  of  the  solution.  This  method 
includes  what  by  many  authors  is  called  ^^Arithmetical  Analy- 
sis.'' 

2.  The  Proportion  Metliod,  in  which  the  solution  is 
obtained  by  forming  and  solving  a  proportion. 

1.     THE  EQUATION  METHOD. 

74.  Arrang-ing  tlie   Parts  of  a  Problem. — For 

the  sake  of  convenience  and  system  in  solution,  the  following 
rules  should  be  observed  in  arranging  the  parts  of  a  problem : 

Rules:  1.  The  question,  or  requirement,  of  a  problem  shoidd  be 
stated  in  the  form  of  an  equation  with  only  the  blank  term  on  the 
right  of  the  sign  of  equality. 

2.  The  basis  of  the  problem  shoidd  be  stated  in  that  form  which 
agrees  in  arrangement  with  the  question,  or  requirement,  as  stated 
according  to  Rule  1. 

Note. — It  is  very  necessary  that  the  pupil  should  learn  to  state  by  the 
above  rules  the  parts  of  a  problem  without  hesitation. 

In  the  following  examples,  the  parts  of  the  problems  are 
properly  arranged  for  solution. 

EXAMPIiES. 

io  If  8  books  cost  $4,  find  the  cost  of  3  books. 
Partsi :    (1)  Cost  of  3  books  =  $(  )  ?    (  Question.) 
(2)  Cost  of  8  books  =  $4.    (  Basis.) 


ADVANCED    ARITHMETIC.  113 

2.  If  8  books  cost  $4,  how  many  books  can  be  bought  for  $10  ? 

Paris  :    (!)  $10  =  cost  of  (  )  books  ?    (Question.) 
(2)  H  =  cost  of  8  books.    ( Basis. ) 

XoTB.— Compare  the  ari*q,ngement  of  parts  in  Nos.  1  and  2, 

3.  Reduce  9  feet  to  inches.     (1  foot=12  inches.) 

Parts:    (1)  9  ft.  =  (  )  in.?    (Question.) 
(2)  1ft.  =  12  in.    (Basis.) 

4.  Reduce  72  inches  to  feet.     (1  inch^j^^j  foot.) 

Parts  :    (1)  72  in.  =  (  )  ft.?    (Question.) 
(2)  1  in.  =  i'2  ft.    (Basis.) 

Note. — Compare  the  arrangement  of  parts  in  Nos.  3  and  4. 

5.  I  owe  a  debt  of  $720,  and  can  pay  45%  of  it.     How  much 
can  I  pay  ? 

Parts  :    ( 1 )  4b%  of  debt  =  $(  )  ?    ( Question . ) 
(2)  100^  of  debt  =  $720.    ( Basis.) 

Note. — Remember  that  100^  of  anything  is  all  of  it. 

6.  I  owe  $120  and  pay  $40.     What  %  of  the  debt  do  I  pay  ? 

Parts  :    (1)  $40  =  (  )%  of  debt  ?    (Question.) 
(2)  $120  =  100^  of  debt.    ( Basis.) 

Note. — Compare  the  arrangement  of  parts  in  Nos.  5  and  6. 

7.  Find  the  interest  on  $500  for  2  years  at  8% .    (Basis  omitted.) 

Parts :    (1)  Int.  on  $500  at  8%  for  2  yr.  =  $(  )  ?    ( Question.) 
(2)  Int.  on  $1  at  1%  for  1  yr.  =  $.01.    ( Basis.) 

8.  Find  the  principal  required  to  produce  $80  interest  in  2 
years  at  8%.     (Basis  omitted.) 

Parts  :    (1)  Prin.  reqd.  to  produce  $80  in  2  yr.  at  8^  =  $(  )  ?    (  Question.) 
(2)  Prin.  reqd.  to  produce  $.01  in  1  yr.  at  1^  =  $1.     (Basis.) 

9.  Find  the  rate  required  for  $500  to  produce  $80  in  2  years. 
(Basis  omitted.) 

Parts :  (1)  Rate  reqd.  for  $500  to  produce  $80  in  2  yr.  =  (  )^  ?    ( Question.) 
(2)  Rate  reqd.  for  $1  to  produce  $.01  in  1  yr.  =  1%.     ( Basis.) 


114  ADVANCED    ARITHMETIC. 

10.  Find  the  time  required  for  $500  to  produce  $80  at  8%. 
(Basis  omitted.) 

Parts  :    (1)  Time  reqd.  for  $500  to  produce  $80  at  8^  =  (  )  yr.?    (Question.) 
(2)  Time  reqd.  for  $1  to  produce  $.01  at  1^  =  1  yr.    (Basis.) 

Note. —  Compare  the  arrangement  of  parts  in  Nos.  7  to  10. 

11.  Find  the  area  of  a  floor  12  feet  long  and  8  feet  wide. 
(Basis  omitted.) 

Parts  :    (1)  Area  of  a  surface  12  ft.  1. ,  8  ft.  w.  =  (  )  sq.  ft.?    ( Question.) 
(2)  Area  of  a  surface  1  ft.  1.,  1  ft.  w.  =  l  sq.  ft.     (Basis.) 

12.  Find  the  length  of  a  floor  12  feet  wide,  containing  240 
square  feet.     (Basis  omitted.) 

Parts  :  (1)  Length  of  a  surface  of  240  sq.  ft.,  12  ft.  w.  =  (  )  ft.?    (Question.) 
(2)  Length  of  a  surface  of  1  sq,  ft.,  1  ft,  w.  =  1  ft.    ( Basis.)     ♦ 

13.  Find  the  width  of  a  surface  of  108  square  feet,  12  feet 
long.     (Basis  omitted.) 

Parts  :    (1)  Width  of  a  surface  of  108  sq.  ft.,  12  ft.  1.  =  (  )  ft.?     (Question.) 
(2)  Width  of  a  surface  of  1  sq.  ft.,  1  ft,  1.  =  1  ft.     ( Basis.) 

Note. — Compare  the  arrangement  of  parts  in  Nos.  11  to  13. 

It  should  be  remembered  that,  in  stating  a  problem,  the 
basis  is  omitted  only  when  it  is  supposed  that  the  pupil  al- 
ready has  the  knowledge  necessary  to  enable  him  to  state  the 
basis  for  himself. 

EXERCISE  LI. 

I.  Arrange  for  solution  the  parts  of  each  problem  in  Exercise 
XLIX. 

II.  Arrange  for  solution  the  parts  of  each  of  the  following  : 

1.  150  ounces  is  |  of  what  number  ? 

2.  Reduce  120  feet  to  inches. 

3.  Reduce  228  inches  to  feet. 

4..  If  5  books  cost  $25,  find  the  cost  of  12  books. 


ADVANCED    ARITHMETIC.  115 

5.  Basis  same  as  in  No.  4.  Find  how  many  oooks  can  be 
bought  for  $60. 

6.  If  a  boy  earns^  $.50  in  5  hours,  what  will  he  earn  in  13 
hours  ? 

7.  Find  how  long  it  will  take  $620  at  6%  to  gain  $9  interest. 
S.  Find  how  much  interest  $740  will  gain  at  8%  in  8^  years. 

9.  Basis  same  as  in  No.  6.  How  long  can  I  hire  him  for 
$8.50? 

10.  Find  the  area  of  a  field  20  rods  long  and  12  rods  wide. 

11.  A  field  of  240  sq.  rd.  is  8  rods  wide.     How  long  is  it  ? 

^  12,  If  5  men  excavate  20  tons  of  dirt  in  2  days,  how  many 
tons  will  12  men  excavate  in  24  days  ? 

IS.  Basis  same  as  in  No.  12.  In  what  time  will  8  men  exca- 
vate 120  tons  ? 

H.  25%  of  my  money  is  $250.     How  much  money  have  I  ? 

15.  Basis  same  as  in  No.  14.  What  per  cent,  of  my  money  is 
$450? 

16.  Find  the  volume  of  a  rectangular  cistern  8  ft.  long,  5  ft. 
wide,  and  7  ft.  deep. 

17.  At  what  rate  will  $800  produce  $20  interest  in  5  months  ? 

18.  What  is  the  time,  when  f  of  the  time  past  noon  is  4  hours  ? 

19.  The  volume  of  a  rectangular  solid  is  860  cu.  ft.,  the 
length  15  ft.,  the  width  8  ft.     Find  the  thickness. 

20.  A  can  do  a  piece  of  work  in  20  days.  How  much  of  it 
can  he  do  in  9  days  ? 

21.  Basis  same  as  in  No.  20,  How  long  will  A  be  in  doing 
-J-J-  of  the  work  ? 

22.  I  bought  an  article  for  $5,  and  sold  it  for  -f  of  the  cost. 
Find  the  selling  price. 

23.  42  is  i  of  what  number  ? 
21^.  400  is  how  many  times  25  ? 


116  ADVANCED    ARITHMETIC. 

75.  Tlie  Process  of  Solving  Problems. — After 
the  parts  of  a  problem  have  been  arranged  according  to  the 
rules  given  in  the  last  article,  the  process  of  solving  by  equa- 
tions consists  of  one  or  both  of  the  following  steps : 

Steps:  1.  Perform  such  operation,  or  operations,  upon  the  basis 
as  will  change  it  to  an  equation,  the  numerical  value  of  whose  first 
member  is  unity. 

2.  Perform  such  operation,  or  operations,  upon  the  equation,  ob- 
tained in  Step  1,  as  will  change  it  to  an  equation  whose  first  mem- 
ber is  identical  with  the  first  member  of  the  question. 

If  the  numerical  value  of  the  first  member  of  the  question  is 
unity,  the  first  step  only  is  used ;  if  the  numerical  value  of  the 
first  member  of  the  basis  is  unity,  the  second  step  only  is  used. 

Example  :  If  24  hats  cost  $12,  how  many  hats  can  be  bought 
for  $16? 

Solution  :    (1)  $16  =  cost  of  (  )  hats  ?    ( Question.) 
(2)  $12  =  cost  of  24  hats.    ( Basis.) 
i^a  of  (2)  =  (3)  $1  =  cos  t  of  2  hats. 
16x(3)  =  (4)  $16  =  cost  of  32  hats,  answer. 

76.  Integral  Solutions. — A  solution  whose  equations 
contain  only  integers  in  their  first  members  is  an  integral  solu- 
tion. 

Note. — Integral  solutions  are  not  always  free  from  fractions  (see  Ex- 
ample 4,  page  117) ;  but  fractions  may  occur  in  the  second  members  of 
the  equations  only. 

EXAMFIiES. 

J.  Find  the  cost  of  10  mats  at  $3  each. 

Solution:    (1)  Cost  of  10  mats  =  ${  )?    (Question.) 
(2)  Cost  of  1  mat  =  $3.    ( Basis. ) 
10x(2)  =  (3)  Cost  of  10  mats  =  $30,  answer. 


ADVANCED    ARITHMETIC.  117 

2.  If  10  mats  cost  $30,  what  will  1  mat  cost  ? 

Solution:    (1)  Cost  of  lmat  =  $(  )?    (Question.) 
(2)  Cost  of  10  mats  =  $30.     ( Basis.) 
^0  of  (2)  =  (3)  Cost  of  1  mat  =  $3,  answer. 

3.  If  10  mats  cost  $30,  what  cost  17  mats  ? 

Solution:    (1)  Cost  of  17  mats  =  $(  )?    (Question.) 
(2)  Cost  of  10  mats  =  $30.    ( Basis. ) 
,V  of  (2)  =  (3)  Cost  of  1  mat  =  $3. 
17x(3)  =  (4)  Cost  of  17  mats  =  $51,  answer. 

4.  If  10  mats  cost  $30,  how  many  can  be  bought  for  $63  ? 

Solution:    (1)  $63  =  cost  of  ()  mats?    (Question.) 
(2)  $30  =  cost  of  10  mats.    ( Basis. ) 
^  of  (2)  =  (3)  $1  =  cost  of  i  mat. 
63x(3)  =  (4)  $63  =  cost  of  *8*  or  21  mats,  answer. 

5.  A  man  has  $75^  and  loans  $35.     What  part  of  his  money 
does  he  loan  ? 

Solution:    (1)  $35  =  (  )  of  his  money?    (Question.) 
(2)  $75  =  all  of  his  money.     (Basis.) 
,»5  of  (2)  =  (3)  $1  =  A  of  his  money. 
35x(3)  =  (4)  $35  =  ?|,  or  {s  of  his  money,  answer. 

6.  12  is  what  part  of  40  ? 

Solution :    (l)12  =  ()of40?    ( Question . ) 
(2)  40  =  all  of  40.    (Basis.) 
^of(2)  =  (3)  1  =  4'^  of  40. 
12  X  (3)  =  (4)  12  =  H ,  or  1^0  of  40,  answer. 

7.  40  is  how  many  times  12  ? 

Solution:    (l)40  =  ()xl2?    (Question.) 
(2)  12  =  1x12,    (Basis.) 
,Vof(2)  =  (3)  1  =  ,',  of  12. 
40x(3)  =  (4)  40=  12,  or  3^x12,  answer. 

8.  If  5  men  can  do  a  piece  of  work  in  42  days,  in  what  time 
can  14  men  do  it  ? 


118  ADVANCED   ARITHMETIC. 

Solution:    (1)  Time  reqd.  by  14  men  =  (  )da.?    (Question.) 
(2)  Time  reqd.  by  5  men  =  42  da.     ( Basis.) 
5x(2)  =  (3)  Time  reqd.  by  1  man  =  210  da. 
^  of  (3)  =  (4)  Time  reqd.  by  14  men  =  15  da.,  answer. 

9.  Find  the  area  of  a  surface  16  rods  long  and  10  rods  wide. 

Solution  :    (1)  Area  of  a  surf.  16  rd.  1.,  10  rd.  w.  =(  )  sq.  rd.?  (  Question.) 
(2)  Area  of  a  surf.  1  rd.  1.,  1  rd.  w.  =  1  sq.  rd.    (Basis.) 
16x(2)  =  (3)  Area  of  a  surf.  16  rd.  1.,  1  rd.  w.  =  16  sq.  rd. 
10x(3)  =  (4)  Area  of  a  surf.  16  rd.  1.,  10  rd.  w.  =  160  sq.  rd.,  answer. 

10.  Find  the  length  of  a  surface  of  160  square  rods,  10  rods 
wide. 

Solution :  (1)  Length  of  a  surf,  of  160  sq.  rd.,  lOrd.w.  =(  )rd.?  (Question.) 

(2)  Length  of  a  surf,  of  1  sq.  rd.,  1  rd.  w.  =  1  rd.    ( Basis.) 
160 X  (2)  =  (3)  Length  of  a  surf .  of  160  sq.  rd.,  1  rd.  w.  =  160rd. 
i^jj  of  (3)  =  (4)  Length  of  a  surf,  of  160  sq.  rd.,  10  rd.  w.  =  16  rd.,  answer. 

11.  Find  the  width  of  a  surface  of  160  square  rods,  16  rods 
long. 

Solution:    (1)  Width  of  a  surf,  of  160  sq.  rd.,  16rd.  1.  =(  )  rd.?  (Question.) 
(2)  Width  of  a  surf .  of  1  sq.  rd.,  1  rd.  l.  =  l  rd.     (Basis.) 
160 X (2)  =  (3)  Width  of  a  surf .  of  160  sq.  rd.,  1  rd.  l.  =  160rd. 
1^6  of  (3)  =  (4)  Width  of  a  surf,  of  160  sq.  rd.,  16  rd.  1.  =  10  rd.,  answer. 

12.  If  5  men  build  100  rods  of  fence  in  10  days,  how  many 
rods  can  8  men  build  in  7  days  ? 

Solution:    (1)  Amt.  built  by  3  men  in  7  da.  =(  )  rd.?    (Question.) 
(2)  Amt.  built  by  5  men  in  10  da.  =  100  rd.    ( Basis.) 
I  of  (2)  =  (3)  Amt.  built  by  1  man  in  10  da.  =  20  rd. 
3^  of  (3)  =  (4)  Amt.  built  by  1  man  in  1  da.  =  2  rd. 
3  X  (4)  =  (5)  Amt.  built  by  3  men  in  1  da.  =  6  rd. 
7x(5)  =  (6)  Amt.  built  by  3  men  in  7  da.  =42  rd.,  answer. 

13.  If  5  men  build  100  rods  of  fence  in  10  days,  how  many 
men  can  build  42  rods  in  7  days  ? 

Solution :    (1)  Force  reqd.  to  build  42  rd.  in  7  da.  =(  )  men  ?    (Question.) 

(2)  Force  reqd.  to  build  100  rd.  in  10  da.  =5  men.    (  Basis.) 
jhi  of  (2)  =  (3)  Force  reqd.  to  build  1  rd.  in  10  da.  =  x^,  or  ^^  man. 


ADVANCED    ARITHMETIC.  119 

10x(3)  =  (4)  Force  reqd.  to  build  1  rd.  in  1  da.  =  1%,  or  I  man. 

42x(4)  =  (5)  Force  reqd.  to  build  42  rd.  in  1  da.  =  21  men. 

J  of  (5)  =  (6)  Force  reqd.  to  build  42  rd.  in  7  da.  =  3  men,  answer. 

14..  If  5  men  can  bviild  100  rods  of  fence  in  10  days,  how 
many  days  will  be  required  for  8  men  to  build  42  rods  ? 

Solution :    (1)  Time  reqd.  for  3  men  to  build  42  rd.  =  (  ) da.?    (Question.) 
(2)  Time  reqd.  for  5  men  to  build  100  rd.  =  10  da.    ( Basis.) 
5  X  (2)  =  (3)  Time  reqd.  for  1  man  to  build  100  rd.  =  50  da. 
lU  of  (3)  =  (4)  Time  reqd.  for  1  man  to  build  1  rd.  =  ^  da. 
I  of  (4)  =  (5)  Time  reqd.  for  3  men  to  build  1  rd.  =  ^  da. 
42x(5)  =  (6)  Time  reqd.  for  3  men  to  build  42  rd.  =  7  days,  answer. 

15.  Reduce  42  gallons  to  quarts. 

Solution:    (1)  42gal.  =  (  )  qt.?    (Question.) 
(2)  lgal.  =  4qt.     (Basis.) 
42  X  (2)  =  (3)  42  gal.  =  168  qt. ,  answer. 

16.  Reduce  172  quarts  to  gallons. 

Solution:    (1)  172  qt.  =  (  )  gal.?    (Question.) 
(2)  1  qt.  =  i  gal.    (Basis.) 
172  X  (2)  =  (3)  172  qt.  =  43  gal . ,  answer. 

17.  Find  the  interest  on  $500  at  6%  per  annum  for  3  years. 

Solution :    (1)  Int.  on  $500  at  Q%  for  3  yr.  =  $(  )  ?    (Question.) 
(2)  Int.  on  $1  at  1%  for  1  yr.  =  .$.01.    (Basis.) 
500  X  (2)  =  (3)  Int.  on  .$500  at  1%  for  1  yr.  =  $5. 
6  X  (3)  =  (4)  Int.  on  $500  at  6%  for  1  yr.  =  $30. 
3  X  (4)  =  (5)  Int.  on  $500  at  6%  for  3  yr.  =  $90,  answer. 

18.  Find  the  rate  required  for  $5(X)  to  gain  $90  in  8  years. 

Solution :    (1)  Rate  reqd.  for  $500  to  gain  $90  in  3  yr.  =  (  )^  ?    (Question.) 

(2)  Rate  reqd.  for  $1  to  gain  $.01  in  1  yr.  =  1%.    (  Basis.) 
sU  of  (2)  =  (3)  Rate  reqd.  for  $500  to  gain  $.01  in  1  yr.  =  ^Jn^. 
9000  X  (3)  =  (4)  Rate  reqd.  for  $500  to  gain  $90  in  1  yr.  =  18^. 
J  of  (4)  =  (5)  Rate  reqd.  for  $500  to  gain  $90  in  3  yr.  =6^,  answer. 


19 
annum 


'.  Find  the  time  required  for  $500  to  gain  $90  at  6%  i^er 


120  ADVANCED   ARITHMETIC. 

Solution :    (1)  Time  reqd.  for  $500  to  gain  $90  at  6$g  =  (  )  yr.?    ((Question.) 

(2)  Time  reqd.  for  $1  to  gain  $.01  at  1^  =  1  yr.    (Basis.) 
gfo  of  (2)  =  (3)  Time  reqd.  for  .$500  to  gain  $.01  at  l%=sU  jr. 
9000 X  (3)  =  (4)  Time  reqd.  for  $500  to  gain  $90  at  1^^  =  18  yr. 
I  of  (4)  =  (5)  Time  reqd.  for  $500  to  gain  $90  at  6^  =  3  yr.,  answer. 

20.  What  principal  will  gain  $90  interest  in  3  years  at  6% 
per  annum  ? 

Solution :    (1)  Prin.  reqd.  to  gain  $90  in  3  yr.  at  6^  =  $(  )  ?    (Question.) 

(2)  Prin.  reqd.  to  gain  $.01  in  1  yr.  at  1%  =  $1.     (Basis.) 
9000 X  (2)  =  (3)  Prin.  reqd.  to  gain  $90  in  1  yr.  at  1^  =  $9000. 
I  of  (3)  =  (4)  Prin.  reqd.  to  gain  $90  in  3  yr.  at  1^  =  $3000. 
i  of  (4)  =  (5)  Prin.  reqd.  to  gain  $90  in  3  yr.  at  6^  =  $500,  answer. 

EXERCISE  LII. 

1.  What  will  20  hats  cost  at  $2.50  each  ? 

2.  21  is  what  part  of  49  ? 

3.  What  cost  22  yards  of  muslin  at  10/  per  yard  ?  18  yards 
of  flannel  at  40/  per  yard  ? 

4..  What  are  40  acres  of  land  worth  at  $50.20  per  acre  ? 

6.  The  light  from  the  sun  reaches  the  earth  in  8  minutes, 
and  it  travels  at  the  rate  of  188000  miles  per  second.  Find 
the  distance  to  the  sun. 

6.  A  has  $840;  B  has  7  times  as  much;  C  has  5  times  as 
much  as  B.     How  much  has  B,  and  C  ? 

7.  What  is  the  area  of  a  surface  120  feet  long  and  30  feet 
wide? 

8.  A  can  travel  on  a  bicycle  4  times  as  fast  as  he  can  walk. 
If  he  can  walk  2  miles  an  hour,  how  fast  can  he  go  on  a  bi- 
cycle ? 

9.  4  acres  is  what  part  of  10  acres  ? 

10.  I  charge  $75  for  collecting  $750 :  what  part  of  the  amount 
collected  will  pay  me  for  collecting  ? 

11.  If  1  man  can  build  a  certain  fence  in  12  days,  how  long 
will  it  take  4  men  to  build  it  ? 


ADVANCED    ARITHMETIC.  121 

12.  An  area  of  140  sq.  ft.  is  7  ft.  wide.     How  long  is  it  ? 

13.  An  area  of  150  sq.  ft.  is  50  ft.  long.     How  wide  is  it  ? 
IJ,..  A's  money,  $500,  is  how  many  times  B's  money,  $125? 

15.  Find  the  volume  of  a  solid  9  ft.  long,  7  ft.  wide,  and  10 
ft.  high.  "    '. 

16.  A  volume  of  630  cu.  ft.  is  9  ft.  long  and  7  ft.  wide.  How 
high  is  it  ? 

17.  A  volume  of  680  cu.  ft.  is  10  ft.  high  and  7  ft.  wide. 
How  long  is  it  ? 

18.  $6800  is  how  many  times  $260  ? 

19.  The  distance  to  the  sun  is  90240000  miles.  Light  trav- 
els from  the  sun  to  the  earth  in  8  minutes :  what  is  the  veloc- 
ity of  light  per  second  ? 

20.  A  lady  earns  $45  per  month  (20  days).  How  much  is 
that  per  day  ? 

21.  If  20  men  harvest  180  acres  in  5  days,  how  much  does 
each  man  harvest  per  day  ? 

22.  Three  leaps  of  a  fox  is  as  far  as  two  leaps  of  a  hound. 
Reduce  60  fox-leaps  to  hound-leaps. 

Question:    60f.  l.  =  (   )  h.  1.? 

23.  Basis  same  as  in  No.  22.  Reduce  60  hound-leaps  to  fox- 
leaps. 

2 J/..  The  minute-hand  of  a  clock  travels  12  times  as  fast  as 
the  hour-hand.  How  long  will  the  hour-hand  be  in  traveling 
20  minute-spaces  ? 

25.  How  long  will  the  minute-hand  be  in  traveling  8  hour- 
spaces  ? 

Question:    Time  reqd.  by  h.  h.  to  travel  20  min.  spaces  =  {  )  min.? 

26.  The  sun  passes  over  15  degrees  of  longitude  in  1  hour : 
how  many  hours  will  it  be  in  passing  over  100  degrees  ? 

27.  If  $4.86  buys  an  exchange  on  London  for  £1,  how  large 
an  exchange  can  I  buy  for  $9720  ? 


122  ADVANCED   ARITHMETIC. 

28.  Find  the  interest  on  $720  for  5  years  at  8%  per  annum. 

29.  What  principal  will  produce  $168  interest  in  4  years  at 
7%  per  annum? 

30.  At  what  rate  will  $540  bear  $81  in  3  years  ? 

77.  Fractional  Solutions. — Fractional  solutions  are 
those  in  which  the  first  members  of  all  equations  are  fractions 
or  are  expressed  as  fractions. 

Preparatory  Step  :  Reduce  the  fractions  in  the  first  members 
of  the  parts  of  the  problem  to  a  common  denominator. 

EXAMFIiES. 

io  I  of  A's  money  is  $4500.     Find  |  of  it. 

Solution :    (1)  Reduced  to  L.  0.  D.,  |  and  t  =  U  and  M. 

(2)  M  of  A's  money  =  $(  )  ?    (Question.) 

(3)  M  of  A's  money  =  $4500.    (  Basis. ) 
-iV  of  (3)  =  (4)  ^V  of  A's  money  =  $300. 

16  X  (4)  =  (5)  M  of  A's  money  =  $4800,  answer. 

2,  If  1  ton  of  hay  costs  $5,  what  costs  -^q  of  a  ton  ? 

Solution:    (1)  Cost  of  t'd  T.  =  $(  )  ?    (Question.) 
(2)  CostofHT.  =  $5.    (Basis.) 
yV  of  (2)  =  (3)  Cost  of  iV  T.  =  $.50. 

7x(3)  =  (4)  Cost  of  1^^  T.  =  $3.50,  answer. 

3.  Find  |  of  $4000. 

Solution:    ( it  f  of  $4000  =  $(  )?  (Question.) 
(2)  I  of  $4000  =  $4000.  (Basis.) 
i  of  (2)  =  (3)  i  of  $4000  =  $500. 

3  X  (3)  =  (4)  I  of  $4000  =  $1500,  answer. 

4-.  342  is  I  of  what  number  ? 

Solution :    (1)  |  of  No.  =  (  )  ?    (Question.) 
(2)  f  of  No.  =  342.     (Basis.) 
Jof  (2)  =  (3)  i  of  No.  =  114. 

4  X  (3)  =  (4)  I  of  No.  =  456,  answer. 


ADVANCED   ARITHMETIC.  128 

5.  What  number  added  to  |  of  itself  will  give  560  ? 

Solution:    (1)  f  of  No.  =  (  )  ?    (Question.) 

(2)  f  of  No.  + 1  of  No.  =560.    (Basis.) 
(2)  =  (a)  Y  of  No.  =  560. 
x^of  (3)  =  (4)  i.of  No.=40. 

9  X  (4)  =  (5)  I  of  No.  =  360,  answer. 

6.  What  number  less  its  f  leaves  510  ? 

Solution :    (1)  ?^  of  No.  =  (  )  ?    (Question. ) 

(2)  I  of  No.  -  f  of  No.  =  510.     ( Basis. ) 

(2)  =  (3)  I  of  No.  =  510. 
iof  (3)  =  (4)  ^  of  No.  =  170. 

7  X  (4)  =  (5)  I  of  No.  =  1190,  answer. 

7.  Find  that  number  whose  |  added  to  its  f  will  give  1240. 

Solution :    (1)  Reduced  to  L.  C.  D.,  |  and  |  =  ^|  and  1%. 

(2)  II  of  No.  =  (  )?    (Question.) 

(3)  \l  of  No.  +  il  of  No.  =  1240.     ( Basis. ) 

(3)  =  (4)  f  i  of  No.  =  1240. 
^Vof  (4)  =  (5)  ^of  No.  =  40. 

24x(5)  =  (6)  iiof  No.  =  960,  answer. 

8.  Find  that  number  whose  ^  less  its  ^  leaves  850. 

Solution :    (1)  Reduced  to  L.  C.  D.,  y'r  and  i  =  ||  and  H. 

(2)  t|of  No.  =  (  )?    (Question.) 

(3)  II  of  No.  -  ii  of  No.  =  850.    ( Basis.) 
(3)  =  (4)  H  of  No.  =850. 

rVof  (4)  =  (5)  ^of  No.  =  50. 
44  X  (5)  =  (6)  It  of  No.  =  2200,  answer. 

D.  720  is  I  more  than  what  number  ? 

Solution:    (1)  fofNo.  =  (  )?    (Question.) 

(2)  I  of  No.  + 1  of  No.  =  720.    ( Basis.) 
(2)  =  (3)  I  of  No.  =  720. 
I  of  (3)  =  (4)  i  of  No.  =  144. 
3  X  (4)  =  (5)  I  of  No.  =  432,  answer. 

10.  102  is  f  less  than  what  number  ? 

Solution:    (1)  |of  No.  =  (  )?    (Question.) 

(2)  I  of  No.  - 1  of  No.  =  102.    ( Basis.) 
(2)  =  (3)  I  of  No.  =  102. 
iof  (3)  =  (4)  i  of  No.  =  34. 
5  X  (4)  =  (5)  I  of  No.  =  170,  answer. 


124  ADVANCED    ARITHMETIC. 

11,  I  is  what  part  of  J  ? 

Solution :    (1)  Reduced  to  L.  0.  D.,  f  and  |  =  f  and  J. 

(2)  f  =  (  )of  J?    (Question.) 

(3)  I  =  all  of  |.     (Basis.) 
iof(3)  =  (4)i=Vof  ^. 

6x(4)  =  (5)  f  =  f  of|. 
.-.  f  isf  of|. 

12.  .12  is  how  many  times  .08  ? 

Solution:    (1)  .12  =  (  )x.08?    (Question.) 
(2)  .08  =  lx.08.    (Basis.) 
iof  (2)  =  (3)  .01  =  i  of  .08. 
12x(3)  =  (4)  .12  =  ^,  or  lJx.08,  answer. 

EXERCISE  LIII. 

1.  I  of  a  number  is  1240.     Find  |  of  the  number. 

2.  My  salary  is  $765.  If  I  spend  |  of  it,  what  amount  do  I 
spend  ? 

S.  A  man  can  build  f  of  a  wall  in  12  days.  In  what  time 
can  he  build  |  of  it  ? 

J/..  In  selling,  a  merchant  lost  y\  of  the  cost  on  an  article 
which  cost  him  $20.24.  How  much  did  he  lose  ?  What  did 
the  article  sell  for  ? 

5.  A  has  ^j  as  much  money  as  B.  If  they  both  have  $6400, 
how  much  has  B  ?     How  much  has  A  ? 

Basis  :  y^  of  B's  money +i}  of  B's  money  =  $6400.    (Why  ?) 

6.  48  is  y%  of  what  number  ? 

7.  i^  of  a  number  is  1542.     Find  the  number. 

8.  3^  of  a  bolt  of  muslin  costs  $5.  What  will  the  whole  bolt 
cost  ? 

9.  1  charge  ^V  of  a  debt  for  collecting  it.  If  I  receive  $240, 
what  was  the  whole  debt  ? 

10.  $950  is  -^-^  more  than  what  number  ? 

11.  $475  is  y\  less  than  what  number  ? 


ADVANCED    ARITHMETIC.  125 

12.  Find  the  number  whose  f  added  to  its  fg-  makes  $920. 

13.  Find  the  number  whose  W  less  its  -^  leaves  $2080. 
IJ^.  What  number  added  to  its  -f^  gives  136  ? 

15.  What  number  legs  its  |  leaves  75  ? 

16.  What  is  I  of  a  ton  of  hay  worth  at  $3.20  per  ton  ? 

17.  If  T^  of  a  farm  is  worth  $1750,  what  is  the  whole  farm 

worth  ? 

18.  B  can  do  |  of  a  piece  of  work  in  f  of  a  day.     What  part 

of  it  can  he  do  in  |  of  a  day  ? 

19.  \\  times  a  number  exceeds  its  y\  by  54.     Find  3^  times 

the  number. 

20.  A  loaned  B  |  of  his  money  and  had  $480  left.     How  much 
money  had  A  before  making  the  loan  ? 

21.  The  sum  of  two  numbers  is  54;  the  smaller  is  \  of  the 
larger.     Find  the  numbers. 

Basis :  f  of  larger  +  h  of  larger  =  54.    (Why  ?) 

78.  Keciprocals  and  Their  Use.— The  Recipro- 
cal of  a  number  is  one  divided  by  that  number. 

EXAMPIiSa 

1.  The  reciprocal  of  16=^. 

Note. — The  reciprocal  of  an  integer  is  expressed  as  a  common  fraction 
whose  numerator  is  1  and  whose  denominator  is  the  given  integer. 

2.  The  reciprocal  of  .03=;^. 

3.  The  reciprocal  of  3.25=^. 

Note. — The  reciprocal  of  a  decimal  is  expressed  as  a  common  fraction 
whose  numerator  is  1  and  whose  denominator  is  the  given  decimal. 

4.  The  reciprocal  of  f=l-7-f=l  Xi= J. 

Note.— The  reciprocal  of  a  common  fraction  is  expressed  by  inverting 
the  fraction. 

5.  The  reciprocal  of  3^,  or  -^=^. 

Note. — Reduce  to  an  improper  fraction  and  proceed  as  in  No.  4  above. 


126  ADVANCED    ARITHMETIC. 

After  stating  the  parts  of  a  problem  for  solution  often  the 
first  step  is  to  proceed  from  what  is  given,  to  unity  (Step  1,  p. 
116).  This  can  always  be  done  directly,  either  in  integral  or 
fractional  problems,  by  employing  the  following  principle: 

Principle  :   The  product  of  a  number  by  its  reciprocal  equals  1 . 

EXAMPIjES. 

1.  If  20  books  cost  $50,  what  cost  1  book? 

Solution :    (1)  Cost  of  1  book  =  $(  )  ?    (Question.) 
(2)  Cost  of  20  books  =  $50.    ( Basis.) 
aVx(2)  =  (3)  Cost  of  1  book  =  $2.50,  answer. 

Note. — Did  we  multiply  by  the  reciprocal  of  20?  We  have  heretofore 
used  "of"  instead  of  "  x  " ;  is  '* of  "  correct  ?    ( See  Article  88.) 

2.  If  8 J  cords  of  wood  are  worth  $8,  what  is  1  cord  worth  ? 

Solution:    (1)  Price  of  1  cord  =  $(  )?    (Question.) 
(2)  Price  of  3^^  cords  =  $8.    ( Basis. ) 
^  of  (2)  =  (3)  Price  of  1  cord  =  $2.40,  answer. 

Note. — 1%  is  the  reciprocal  of  Sh 

3.  At  $1.25  per  bu.,  how  much  wheat  can  be  bought  for 

$22.50  ? 

Solution:    (1)  $22.50  =  cost  of  ()  bu.?    (Question.) 
(2)  $1.25  =  cost  of  1  bu.    ( Basis. ) 

j^x(2)  =  (8)  $l  =  cost  of^bu. 
22.50  X  (3)  =  (4)  $22.50  =  cost  of  f^,  or  18  bu. ,  answer. 
Note. —  j-^g  is  the  reciprocal  of  1.25. 

4.  12i  is  what  part  of  37i  ? 

Solution:    (1)  12|  =  (  )  of  371^?    (Question.) 
(2)  37i  =  allof37i.     (Basis.) 
^of(2)  =  (3)  l  =  ^of37|. 

121  X  (3)  =  (4)  12i^  =  ^  X  ^ ,  or  ^  of  371,  answer. 
3 


ADVANCED   ARITHMETIC.  127 

EXERCISE  LIV. 

Solve  the  problems  in  Exercise  LIII  by  using  reciprocals  instead 
of  the  fractional  solution  there  given.     Thus, 

1.  Solution:    (1)  |'of  No.  =  (  )?    (Question.) 

(2)  I  of  No.  =  1240.     (Basis.) 
f  of  (2)  =  (3)  Ix  No.  =  1550. 
I  of  (3)  =  (4)  I  of  No.  =  1356i,  answer. 

79.  Solutions  Shortened. — While  it  is  strenuously 
urged  that  the  pupil  should  always  follow  out  a  complete  solu- 
tion for  every  problem,  it  will  become  monotonous  and  even 
repulsive  to  the  pupil  to  be  compelled  always  to  go  the  "  long 
way  around,"  when  he  can  grasp  completely  a  more  direct  so- 
lution. 

Since  solving  a  problem  consists  in  commencing  with  a  known 
condition  and  passing  by  means  of  a  known  authorized  process 
from  that  known  condition  to  the  unknown  (required)  condition^ 
then  it  follows  that  both  the  known  condition  and  the  process 
employed  may  depend  largely  upon  the  knowledge  and  ability 
of  the  person  taught.  The  following  suggestions  will  aid  in 
shortening  many  solutions : 

(1)  Common  fractions  are  usually  to  be  preferred  to  deci- 
mals.    Learn  the  following  table : 

EaUIVAIiENTS. 

f.06y4=r/i6 
f. 081/3= $yi2 

$.10     =$Vio 

$.i2y2=r/8 
$.i6%=r/6 
$.20  =r/5 

$.25    =$1/4 

$.331/3  =  $l/3 

$.50    =$1/2 


128  ADVANCED    ARITHMETIC. 

EXAMPIiES. 

1.  At  16f/  each,  how  many  knives  can  be  bought  for  $8.50  ? 

Solution:    (1)  $3|  =cost  of  (  )  knives?    (Question.) 
(2)  $^  =  cost  of  1  knife.    ( Basis. ) 
6x(2)  =  (3)  $l  =  cost  of  6  knives. 
31  x(3)  =  (4)  $3J  =  cost  of  21  knives,  answer. 

2.  What  will  220  dozen  eggs  sell  for,  at  12^/  per  dozen  ? 

Solution :    (1)  Price  of  220  doz.  =  $(  )  ?    (Question.) 
(2)  Price  of  1  doz .  =  $| .    ( Basis . ) 
220 X (2)  =  (3)  Price  of  220  doz.  =  $27^,  or  $27.50,  answer. 

(2)  If  the  first  member  of  the  basis  is  a  known  factor  of  the 
first  member  of  the  question,  one  step  only  is  needed  in  solving 
the  problem. 

EXAMPIiES. 

1.  4  times  A's  money  is  $800.  If  B  has  12  times  as  much 
money  as  A,  how  much  has  B  ? 

Solution:    (1)  12 x A's  money  =  $(  )?    (Question.) 
(2)  4  X  A's  money  =  $800.    ( Basis. ) 
3  X  (2)  =  (3)  12  X  A's  money  =  $2400,  answer. 

2.  Reduce  128  ounces  avoirdupois  to  pounds. 

Solution:    (1)  128  oz.  =  (  )  lb.  ?    (Question.) 
(2)  16oz.  =  llb.    (Basis.) 
8  X  (2)  =  (3)  128  oz.  =  8  lb. ,  answer. 

Note.— Here  the  basis  of  solution  is  not  given,  and  may  be  stated  in 
either  of  the  forms :  16  oz.  =  1  lb.,  or  1  oz.  =  i\  lb.  As  the  number  to  be 
reduced  (128  oz.)  is  a  multiple  of  16  oz.,  the  first  form  is  as  good  as  the 
second,  if  not  better.  But  if  the  number  to  be  reduced  had  not  been  a 
multiple  of  16  oz.,  the  second  form  would  have  been  the  better,  since 
it  starts  with  unity  in  its  first  member. 

(3)  If  the  first  member  of  the  basis  is  a  known  multiple  of 
the  first  member  of  the  question,  one  step  only  is  needed  in 
solving  the  problem. 


ADVANCED    ARITHMETIC.  129 

EXAMPLES. 

1.  If  $85  will  buy  30  books,  how  many  books  can  be  bought 
for  $17? 

Solution:    (1)  $17'=f  price  of  ()  books ?    (Question.) 
(2)  $85  =  price  of  30  books.    (Basis.) 
\  of  (2)  =  (3)  $17  =  price  of  6  books,  answer. 

2.  If  5  men  can  do  a  piece  of  work  in  45  days,  in  what  time 
can  15  men  do  the  same  work  ? 

Solution:    (1)  Time  reqd.  by  15  men  =  (  )  da.?    (Question.) 
(2)  Time  reqd .  by  5  men  =  45  da.    ( Basis. ) 
I  of  (2)  =  (3)  Time  reqd.  by  15  men  =  15  da.,  answer. 

3.  A  bought  two  tracts  of  land  at  a  uniform  price  per  acre. 
The  first  was  80  rods  long  and  40  rods  wide,  and  cost  $4000. 
What  did  the  other  cost,  which  was  20  rods  long  and  8  rods 
wide  ? 

Solution  :    (1)  Cost  of  a  tract  20  rd.  1.,  8  rd.  w.  =$(  )  ?    (Question.) 
(2)  Cost  of  a  tract  80  rd.  1.,  40  rd.  w.  =$4000.    (Basis.) 
iof  (2)  =  (3)  Costof  atract20rd.l.,40rd.  w.  =  $1000. 
\  of  (3)  =  (4)  Cost  of  a  tract  20  rd.  1.,  8  rd.  w.  =  $200,  answer. 

(4)  When  the  basis  of  a  problem  is  omitted,  the  pupil 
should  be  permitted  to  use  whatever  basis  he  can  readily  un- 
derstand. 

EXAMPIiE. 

Find  the  interest  on  $5(X)  for  2^  years  at  6%  per  annum. 

Solution  :    (1)  Int.  on  $500  at  Q%  for  21  yr.  =  $(   )  ?    (Question.) 
(2)  Int.  on  $500  at  Q%  for  1  yr.  =  $30.     (  Basis.) 
2^x(2)  =  (3)  Int.  on  $500  at  6^  for  2^  yr.  =  $75,  answer. 

Note. — Most  advanced  pupils  will  readily  see  that  the  ^^ Interest  on 
$500  for  1  yr.  at  6%  =  $30'' ;  others  may  be  able  only  to  see  that  ''Interest 
on  $500  for  1  yr.  at  1%  =  $5''  ;  while  the  beginner  will  have  to  commence 
with  ''Interest  on  $1  for  1  yr.  at  1%  =  $.01." 


130  ADVANCED    ARITHMETIC. 

EXERCISE  LV. 

i.  What  will  50  lb.  of  butter  cost,  at  20/  per  pound  ? 

2,  What  are  80  knives  worth,  at  25/  each  ? 

3,  I  bought  15  bu.  of  corn,  at  83^/  per  bu.  How  much  did 
it  cost  me  ? 

If..  What  will  64  yds.  of  cloth  cost,  at  6^/  per  yd.? 

5.  I  bought  24  quarts  of  strawberries,  at  8 J/  per  qt.  How 
much  did  I  pay  for  them  ? 

6.  At  12-^/  each,  how  many  tin  pails  can  be  bought  for  $4  ? 

7.  At  25/  each,  how  many  melons  can  be  bought  for  $5  ? 

8.  At  50/  per  bu.,  how  many  bushels  of  apples  can  be  bought 
for  $12  ? 

9.  At  33J/  each,  how  many  books  can  be  bought  for  $11  ? 

10.  15  is  what  part  of  75  ? 

11 .  A  bill  of  exchange  on  London  costs  $4.86  per  £.  How 
many  £'s  will  the  bill  call  for,  if  it  cost  $486  ? 

Note. — The  English  pound,  or  sovereign,  £,  is  worth  about  $4.86  in 
American  gold. 

12.  Reduce  315  gallons  to  barrels  (81.5  gal.  =  l  bbl.). 

13.  A  block  of  marble  10  ft.  long,  8  ft.  wide,  and  6  ft.  high, 
is  worth  $420.  Find  the  value  of  a  block  5  ft.  long,  2  ft.  wide, 
and  2  ft.  high. 

IJf..  A  has  $1000,  which  is  3  times  as  much  as  B  has ;  C  has  9 
times  as  much  as  B.     How  much  has  C  ? 

15.  If  20  men  build  24  miles  of  fence  in  8  days,  in  how  many 
days  can  5  men  build  8  miles  ? 

16.  Since  80°  Reaumur  is  as  hot  as  100°  Centigrade,  how 
many  degrees  Centigrade  will  correspond  to  240°  Reaumur  ? 

17.  Reduce  160  ounces  avoirdupois  to  pounds. 

18.  The  sun  passes  over  15  degrees  of  longitude  in  1  hour. 
In  what  time  will  it  pass  over  185  degrees  ? 

19.  John,  who  is  16  years  old,  is  4  times  as  old  as  James; 
Henry  is  12  times  as  old  as  James :  how  old  is  Henry  ? 


ADVANCED    ARITHMETIC.  181 

20.  If  96  pounds  of  sea-water  contains  3  pounds  of  salt,  how 
many  pounds  of  salt  in  16  pounds  of  sea-water  ? 

21.  If  the  interest  on  $500  for  a  certain  time  at  a  certain 
rate  is  $25,  what  is  the  interest  on  $125  for  the  same  time  at 
the  same  rate  ? 

22.  15  times  A's  money  is  $720.  How  much  has  B,  who  has 
5  times  as  much  as  A  ? 

23.  The  wages  of  3  men  for  2  days  is  $9.  Find  the  wages  of 
12  men  for  8  days. 

2Jf..  If  20  men  in  15  days  build  5  miles  of  fence,  how  many 
miles  can  5  men  build  in  45  days  ? 

80.  Percentage  Solutions. — In  arithmetic,  the  term 
Per  Cent  means  liundredtli  or  liundredtlis.    The 

sign,  %,  is  used  for  the  words  per  cent. 

l%=Tb; 

12i%=f|*;  and 
100%=fft«. 

You  should  not  think  of  percentage  as  something  new  in 
principle ;  it  is  only  a  new  form  of  expressing  hundredths. 

Principle:  100%  of  any  number  is  all  of  it. 

To  aid  in  shortening  percentage  solutions  the  following  ta- 
bles are  given,  and  should  be  committed  to  memory: 


2%=y5o  of  100%  I2y2%=y8  of  100% 

4%=y25  of  100%  i6%%=y6  of  100% 

5%=y2oof  100%  20%=y5of  100% 

6y4%=i/i6  of  100%  25%=y4  of  100% 

81/3%  =1/12  of  100%  33y3%=:y3  of  100% 

10%-yio  of  100%  50%=y2  of  100#7 


.   UNIVERSITY 


182  ADVANCED   ARITHMETIC. 

100%  =  50x2%  -100%  =8x121/2% 
100% =25x4%  100% =6x16%% 

100% =20x5%  100% =5x20% 

100%  =  16x6V4%  100%=4x25% 

100%  =  12x81/3%  100%  =3x33%% 

100%  =  10xlO%  100%=2x50% 

i.  Find  12%  of  630  pounds. 

Solution :    (1)  12^  of  630  lb.  =  (  )  lb.?    (Question.; 

(2)  IW  of  630  lb.  =  630  lb.     (Basis.) 
T^  of  (2)  =  (3)  1%  of  630  lb.  =  6.3  lb. 

12  X  (3)  =  (4)  12^  of  630  lb.  =  75.6  lb. ,  answer. 

2.  Mr.  Jones  has  $1600,  of  which  25%  is  in  the  bank.  How 
much  has  he  in  the  bank  ? 

Solution:    (1)  25^  of  .$1600  =  (  )?    (Question.) 
(2)  100^  of  $1600  =  $1600.     (Basis.) 
i  of  (2)  =  (3)  25^  of  $1600  =  $400,  answer. 

Note. — By  knowing  (from  the  table)  that  25%  is  I  of  100^,  we  at  once 
take  I  of  the  basis,  which  gives  the  required  answer  without  having  to 
pass  first  to  unity. 

3,  I  sell  goods  for  Mr.  Carr,  amounting  to  $400,  on  which  I 
charge  a  commission  of  2%.     Find  my  commission. 

Solution:    (1)  2^  of  sales  =  $(  ) ?    (Question.) 
(2)  100^  of  sales  =  $400.     ( Basis.) 
6»5  of  (2)  =  (3)  2%  of  sales  =  $8,  answer. 

Note. — Commission  for  selling  is  always  some  %  of  the  selling  price. 

4..  I  bought  goods  for  a  merchant,  amounting  to  $80.  I 
charged  5%  commission.     Find  my  commission. 

Solution:    (1)  5^  of  purchase  =  $()  ?    (Question.) 
(2)  100^  of  purchase  =  $80.     (  Basis. ) 
^  of  (2)  =  (3)  5%  of  purchase  =  $4,  answer. 

Note. — Commission  for  buying  is  always  some  %  of  the  purchase  price. 


ADVANCED    ARITHMETIC.  188 

5.  An  article  cost  $4.20,  and  was  sold  at  a  profit  of  16f  %. 
Find  the  gain. 

Solution:    (1)  .16|^  of  cost  =  $(  )?    (Question.) 
(2)  100^  of  cost  =  $4.20.     ( Basis.) 
i  of  (2)  =  (3)  161^  of  cost  =  $.70,  answer. 

•    Note. — The  profit  or  loss  is  always  some  %  of  the  cost  price. 

6.  I  bought  goods  for  $800,  and  sold  them  at  88^%  profit. 
Find  the  selling  price. 

Solution  :    (1)  33 J^  of  cost  =  $(  )  ?    (Question.) 
(2)  100^  of  cost  =  $300.     (Basis.) 
I  of  (2)  =  (3)  33J^  of  cost  =  $100. 

(4)  $300+$100  =  $400,  answer. 

Note. — The  profit  added  to  the  cost  gives  the  selling  price. 

7.  I  bought  a  horse  for  $72,*  and  sold  him  at  a  loss  of  12^%. 
Find  the  selling  price. 

Sohition:    (1)  12|^  of  cost  =  $(  )?    (Question.) 
(2)  100^  of  cost  =  $72.     (  Basis. ) 
i  of  (2)  =  (3)  121^  of  cost  =  $9. 

(4)  $72 -$9  =  $63,  answer. 

Note. — The  loss  subtracted  from  the  cost  gives  the  selling  price. 

8.  VI %  of  a  number  is  68.     Find  the  number. 

Solution:    (1)  100^  of  number  =  (  )?    (Question.) 
(2)  17^  of  number  =  68.    (Basis.) 
yJ,  of  (2)  =  (3)  1%  of  number  =  4. 
100  X  (3)  =  (4)  100^  of  number  =  400,  answer. 

Note. — When  we  cannot  solve  by  a  shorter  plan,  we  can  always  go 
from  what  we  have  given  to  1%;  then  from  1%  to  the  required  num- 
ber of  %. 

9.  A  lawyer  is  paid  5%  for  collecting  a  debt.  If  he  gets  $45, 
what  is  the  amount  of  the  debt  ? 

Solution:    (1)  100^  of  debt  =  $(  )  ?    (Question.) 
(2)  b%  of  debt  =  $45.     ( Basis. ) 
20  X  (2)  =  (3)  100^  of  debt  =  $900,  answer. 


184  ADVANCED    ARITHMETIC. 

10.  My  gain  on  a  certain  article,  which  I  sold  at  a  profit  of 

25%,  is  $.75.     Find  the  cost  price. 

Solution:    (1)  100^  of  cost  =  $(  )?    (Question.) 
(2)  25^  of  cost  =  $.75.     (Basis.) 
4x(2)  =  (3)  100^  of  cost  =  $3,  answer. 

11.  My  loss  on  a  certain  article,  which  I  sold  at  a  loss  of 
16|%,  is  $1.25.     Find  the  cost  price. 

Solution:    (1)  KW  of  cost  =  $(  )?    (Question.) 
(2)  161^  of  cost  =  $1.25.    (Basis.) 
6  X  (2)  =  (3)  IW  of  cost  =  $7.50,  answer. 

12.  In  one  book-case  I  have  50  books ;  this  is  12-^%  of  my 
library.     How  many  books  have  I  in  my  library  ? 

Solution:    (1)  100^  of  library  =  (  )  books  ?    (Question.) 
(2)  121  ^  of  library  =  50  books.     ( Basis. ) 
8  X  (2)  =  (3)  100^  of  library  =  400  books,  answer. 

13.  55  is  25%  more  than  what  number  ? 

Solution  :    (1)  100^  of  No.  =  (  )  ?    (Question.) 

(2)  100^  of  No. +  25^  of  No.  =  55.    (  Basis.) 
(2)  =  (3)  125^  of  No.  =55. 
^of  (3)  =  (4)  25^  of  No.  =  11. 
4  X  (4)  =  (5)  100^  of  No.  =  44,  answer. 

14..  An  article  was  sold  at  50%  above  cost  for  $4.20.     What 
did  the  article  cost  ? 

Solution  :    (1)  100^  of  cost  =  $(  )  ?    (Question.) 

(2)  lOO^of  cost +50^  of  cost  =  $4.20.    (Basis.) 
(2)  =  (3)  150^  of  cost  =  $4.20. 
^  of  (3)  =  (4)  50^  of  cost  =  $1.40. 
2  X  (4)  =  (5)  100^  of  cost  =  $2.80,  answer. 

15.  70  is  12^%  less  than  what  number  ? 

Solution :    (1)  100^  of  No.  =  (  )  ?    (Question.) 

(2)  100^  of  No.  -12|^  of  No.  =  70.     (Basis.) 
(2)  =  (3)  871^  of  No.  =  70. 
i.of  (3)  =  (4)  121^  of  No.  =  10. 
8  X  (4)  =  (5)  100^  of  No.  =  80,  answer. 


ADVANCED    ARITHMETIC.  135 

16.  $7  is  what  %  of  $20  ? 

Solution:    (1)  $?  =  (  )^  of  $20?    (Question.) 
(2)  $20=  100^  of  $20.    (Basis.) 
aVof  (2)  =  (3)  $l  =  5^of  $20. 

7  X  (3)  =  (4)  $7  =  35^  of  $20,  answer. 

17.  A.  lawyer  gets  $85  commission  for  collecting  a  debt  of 
$500.     What  %  commission  does  he  charge  ? 

Solution  :    (1)  $35  =  (  )^  of  debt  ?    (Question. ) 

(2)  $500  =  100^  of  debt.    ( Basis. ) 
5^^  of  (2)  =  (3)  $l  =  k^ofdebto 
35  X  (3)=(4)  $35  =  1%  of  debt,  answer. 

18.  I  paid  $52  for  a  buggy,  and  sold  it  so  as  to  gain  $13.     Find 
the  %  of  profit. 

Note. — As  profit  or  loss  is  some  per  cent  of  the  cost,  in  this  problem 
we  really  want  to  know  how  many  %  $13  is  of  the  cost,  or  $52. 

Solution:    (1)  $13  =  (  )^  of  cost ?    (Question.) 
(2)  $52  =  100^  of  cost.     (Basis . ) 
i  of  (2)  =  (3)  $13  =  25^  of  cost,  answer. 

19.  Find  8%  of  15%  of  $3500. 

Solution  :    (1)  15^  of  $3500  =  $(  )  ?  (Question.) 

(2)  100^  of  100^  of  $3500  =  $3500.  (Basis.) 
I  i^  of  (2)  =  (3)  1%  of  $3500  =  $35. 
15  X  (3)  =  (4)  15^  of  $3500  =  $525. 

(5)  8%  of  $525  =  $(  )  ?  (Question.) 

(6)  100^  of  $525  =  $525.     ( Basis. ) 
r^TT  of  (6)  =  (7)  1%  of  $525  =  $5.25. 

8  X  (7)  =  (8)  8%  of  $525  =  $42,  answer. 

Plan.— The  plan  of  this  solution  is  to  find  15%  of  $3500;  then,  8%  of 
that  result.    This  gives  8%  of  15^  of  $3500. 

Another  form:    (1)  8^  of  15^  of  $3500  =  $(  )?    (Question.) 
(2)  100^  of  $3500  =  $3500.    ( Basis. ) 
jU  of  (2)  =  (3)  1%  of  100^  of  $3500  =  $35. 
jU  of  (3)  =  (4)  1%  of  1%  of  $3500  =  $.35. 
8  X  (4)  =  (5)  8^  of  1%  of  $3500  =  $2.80. 
15  X  (5)  =  (6)  8%  of  lb%  of  $3500  =  $42,  answer. 


136  ADVANCED    ARITHMETIC. 

20.  75  is  how  many  %  more  than  60  ? 

Solution :    (1)  75  =  (  )^  of  60  ?    (Question. ) 
(2)  60  =  100^  of  60.     (Basis.) 
JffOf(2)  =  (3)  1  =  -V^of60. 
75x(3)  =  (4)  75  =  125^  of  60. 

.-.  75  is  25^  more  than  60. 

Note.—  Since  100^  of  60  is  all  of  60,  125^  of  60  is  25^  more  than  60. 


21.  40  is  how  many  %  less  than  50  ? 

Solution:    (1)  40  =  (  )^  of  50?    (Question.) 
(2)  50  =  100^  of  50.     (Basis,) 
sVof  (2)  =  (3)  1  =  2^  of  50. 
40x(3)  =  (4)  40  =  80^  of  50. 

.-.  40  is  20^  less  than  50. 

22.  A's  money  is   25%  more  than  B's;  then,  B's  money  is 
how  many  %  less  than  A's  ? 

Solution :    (1)  100^  of  B's  money  =  (  )^  of  A's  money  ?    (Question.) 

(2)  100^  of  B's  money  +  25^  of  B's  money  =  100^  of  A's  money. 

(2)  =  (3)  125^  of  B's  money  =  100^  of  A's  money. 
T^  of  (3)  =  (4)  1%  of  B's  money  =  i%  of  A's  money. 
100x(4)  =  (5)  100^  of  B's  money  =  80^  of  A's  money. 

.'.  B's  money  is  20^  less  than  A's. 

23.  A's  money  is  20%  less  than  B's;  then,  B's  money  is  how 
many  %  more  than  A's  ? 

Solution  :    (1)  100^  of  B's  money  =  (  )^  of  A's  money  ?    (Question.) 

(2)  100^  of  B's  money  -  20^  of  B's  money  =  100^  of  A's  money. 

( Basis.) 

(2)  =  (3)  80^  of  B's  money  =  100^  of  A's  money, 
i  of  (3)  =  (4)  20^  of  B's  money  =  25^  of  A's  money. 
5x(4)  =  (5)  100^  of  B's  money  =  125^  of  A's  money. 

.'.  B's  money  is  25^  more  than  A's. 


ADVANCED    ARITHMETIC.  137 

EXERCISE  LVI. 

1.  Find  6%  of  $800. 

2.  Find  12i%  of  $4000. 

3.  Find  20%  of  i  of  $1600. 

^.  8%  of  80%  of  $860  is  how  much  ? 
5.  Find  40%  of  60%  of  $i. 

^.  I  owe  A  $640;    B,  $880;   C,  $720;  D,  $.25.     How  much 
will  each  get,  if  I  can  pay  but  20%,  or  $.20  on  the  $1  ? 

7.  A  certain  tax  levy  is  1^%.     Find  A's  taxes,  if  his  prop- 
erty is  worth  $46420. 

8.  9%  of  a  debt  is  $189.72.     What  is  the  debt  ? 

9.  $3200  is  16%  of  what? 

10.  $3200  is  40%  of  what  ? 

11.  $75  is  88^%  more  than  what  number  ? 

12.  $80  is  16|%  less  than  what  number  ? 

13.  $250  is  how  many  %  more  than  $200  ? 
IJ^.  $750  is  how  many  %  less  than  $900  ? 

15.  A's  money  is  12|%  more  than  B's;  then,  B's  money  is 
how  many  %  less  than  A's  ? 

16.  A's  money  is  25%  less  than  B's ;  then,  B's  money  is  how 
many  %  more  than  A's  ? 

17.  A  has  a  flock  of  1800  sheep.  He.  sells  15%  to  B,  who 
sells  10%  of  his  to  C.     After  these  sales  how  many  has  each  ? 

18.  I  drew  from  a  bank  $60,  which  was  15%  of  my  deposit. 
What  sum  had  I  left  in  the  bank  ? 

19.  B  sells  a  pair  of  shoes  at  10%  less  than  list  price;  if  list 
price  is  38^%  above  cost,  and  B  sells  them  for  $1.80,  find  the 
cost. 

20.  What  is  the  face  value  of  Government  bonds  which  imy, 
at  6%,  an  income  of  $720  per  annum  ? 

21.  1  bushel  is  how  many  per  cent  more  than  3  pecks  ? 

22.  1  yard  is  how  many  per  cent  less  than  1  meter  ? 
Note.— 36  inches  =  1  yard;  39.37  inches  =  1  meter. 

23.  1  meter  is  how  many  per  cent  more  than  1  yard  ? 


188  ADVANCED    ARITHMETIC. 

^4.   1  pint  is  what  per  cent  of  1  gallon  ? 

25.  5  months  is  what  %  of  9  months  ? 

26.  20%  of  50%  of  $600  is  what  per  cent  of  80%  of  30%  of 
$400? 

Plan.— (1)  Find  80^  of  3(¥  of  $400;  (2)  find  20^  of  60^  of  $600 ;  (3)  find 
what  %  the  second  result  is  of  the  first  result. 

Study  the  definitions  and  principles  (omitting  the  formulas 
and  relations),  and  solve  by  the  method  here  given  all  the 
problems  in  Profit  and  Loss,  Trade  Discount,  Commission, 
Stocks  and  Bonds,  Taxes,  Duties,  and  Insurance.  (  See  Part 
II,  pp.  251-280.) 

3.     THE  PROPORTION  METHOD. 

81.  Process. — The  process  of  solving  a  problem  by  the 
'proportion  method  consists  in  (1)  stating  a  proportion  according 
to  the  following  principle,  and  (2)  solving  that  proportion. 

Principle:  In  two  equations  of  the  same  nature^  and  depending 
upon  the  same  condition,  the  numerical  ratio  between  the  first  mem- 
bers is  equal  to  the  numerical  ratio  between  the  second  members. 

HiliTTSTRATION. 

Suppose,  (a)  Cost  of  1  book =$3, 

Then,  40  x  (1)  =  (b)  Cost  of  40  books  =  $120. 

30  X  (1)  =  (c)  Cost  of  30  books  =  $90. 

20  X  (1)  =  (d)  Cost  of  20  books  =  $60. 

15  X  (1)  =  (e)  Cost  of  15  books  =  $45. 

5  X  (1)  =  (f )  Cost  of  5  books  =  $15. 

Note. — The  last  five  equations  were  all  obtained  from  the  first,  and 
(1)  are  all  of  the  same  nature,  having  ''cost  of  books"  for  their  first  mem- 
bers and  "^'s"  for  their  second  members;  (2)  they  all  depend  upon  the 
same  condition — that  1  book  costs  $3. 

1.  Compare  (b)  (d). 

(b)  Cost  of  40  books  =  $120. 
(d)  Cost  of  20  books  =  $60. 

(1)  40:20  =  2. 

(2)  120:60  =  2. 

.-.  40: 20::  120: 60. 


ADVANCED    ARITHMETIC.  189 

2.  Compare  (c)  and  (f). 

(c)  Cost  of  30  books  =  $90. 
(f)  Cost  of  5  books  =  $15. 

(1)  30:5  =  6. 

(2)  90:15  =  6.    \ 
.-.  15:5::90:15. 

3.  Compare  (e)  and  (f). 

(e)  Cost  of  15  books  =  $45. 

(f)  Cost  of  5  books  =  $15. 
(1)  15:5  =  3. 
(2)45:15  =  3. 

/.  15:5::45:15. 

When  the  parts  of  a  problem  are  stated  for  solution,  the 
equations  formed  are  of  the  same  nature  and  depend  upon  the 
same  condition ;  therefore,  the  above  principle  may  be  used  in 
stating  the  proportion.  After  the  proportion  is  stated  it  is 
solved  by  methods  in  Articles  68  and  69. 

EXAMPLES. 

1.  If  10  books  cost  $20,  what  will  85  books  cost? 

Solution :    (1)  Cost  of  35  books  =  $(  )  ?    (Question. ) 

(2)  Cost  of  10  books  =  $20.    ( Basis.) 

(3)  35 :  10 : :  (  ) :  20  ?    ( Proportion. ) 

(4)?#=70. 

.*.  the  required  number  is  $70. 

2.  If  I  can  buy  15  hogs  for  $75,  how  many  can  I  buy  for  $90  ? 

Solution  :    (1)  $90  =  cost  of  (  )  hogs  ?    (Question.) 
(2)  $75  =  cost  of  15  hogs.     ( Basis. ) 
(3)90:75::(  ):15?    (Proportion.) 
18 

.'.  the  required  number  is  18  hogs. 


140  ADVANCED    ARITHMETIC. 

3.  Reduce  8  bushels  to  pecks. 

Solution  :    (1)  8  bu.  =  (  )  pk.?    (Question.) 

(2)  lbu.=4pk.    (Basis.) 

(3)  8:1::(  ):4?    (Proportion.) 

(4)  ^  =  32. 

.*.  the  required  number  is  32  pecks. 

4-.  If  30  acres  of  laud  cost  $450,  how  much  was  that  per  acre  ? 

Solution  :    (1)  Cost  of  1  acre  =  $(  )  ?    (Question.) 

(2)  Cost  of  30  acres  =  $450.     ( Basis. ) 

(3)  1:30::(  ):450?    (Proportion.) 

(4)i#=15. 

.*.  the  required  answer  is  $15. 

5.  If  50  shares  of  stock  cost  $4000,  what  will  78  shares  cost  ? 

Solution:    (!)  Price  of  78  shares  =  $(  )?    (Question.) 

(2)  Price  of  50  shares  =  $4000.    ( Basis.) 

(3)  78 :  50 : :  (  ) :  4000  ?    ( Proportion. ) 

(4)18^  =  6240. 
oO 

.-.78  shares  will  cost  $6240. 

6.  A  tract  of  land  40  rods  long,  20  rods  wide,  contains  5 
acres.  What  must  be  the  length  of  a  tract  60  rods  wide  to  con- 
tain 80  acres  ? 

1st  Solution:  (1)  Length  of  a  tract  of  30  A.,  60rd.  w.  =  (  )rd.?    (Question.) 
(2)  Length  of  a  tract  of  5  A. ,  20  rd.  w.  =  40  rd.     ( Basis. ) 

(3)g:^::(  ):40?    (Proportion.) 

.*.  the  required  length  is  80  rods. 

"30  "  5 

Note. —    —  „  and     ^,,  do  not  in  fact  truly  represent  the  numerical 

values  of  the  two  lengths  considered  in  this  problem.    The  "30"  and 


ADVANCED   ARITHMETIC.  141 

"  5  "  are  acresj  while  the  "  60  "  and  "  20  "  are  rods.    There  are  160  square 

J    •  ,30x160       -5x160         ,,  ,.  4.1,    ^ 

rods  in  an  acre,  and  — ^t. —  and  — ^^ —  would  represent  the  true  numer- 
oO  20 

ical  length.    But,  in  working  out  the  proportion,  one  of  these  160's 
would  cancel  the  other,  and  they  may  therefore  be  omitted  altogether. 

It  is  not  necessary  in  the  proportion  method  that  the  blank 
term  should  always  appear  alone  in  the  fight  member  of  the 
question,  or  in  the  right  member  at  all  for  that  matter ;  it  may 
fall  anywhere  in  the  equation,  but  care  must  always  be  taken 
to  arrange  the  question  and  basis  in  the  same  order.  Number  6 
may  be  solved  as  follows : 

2d  Solution :  (1)  Area  of  a  tract  (  )  rd.  1.,  60  rd.  w.  =  30  A.?    (Question.) 
(2)  Area  of  a  tract  40  rd.  L,  20  rd.  w.  =  5  A.    (Basis.) 
(8)  (  )x60:40x20::30:5?    (Proportion.) 

40X20X30 
^  '       60x5 
.•.  the  required  length  is  80  rods. 

7.  A  tract  of  land  40  rods  long,  20  rods  wide,  is  worth  $8(X). 
Find  the  width  of  a  tract  50  rods  long  that  is  worth  $1500. 

Solution  :    (1)  Value  of  a  tract  50  rd.  1.,  (  )  rd.  w.  =  $1500 ?    (Question.) 

(2)  Value  of  a  tract  40  rd.  1.,  20  rd.  w.  =  $800.     ( Basis.) 

(3)  50  X  (  ) :  40  X  20 : :  1500 :  800  ?    ( Proportion ,) 
.  .  40x20x1500 _ 

^^^       50X800      "^^- 

.*.  the  required  width  is  30  rods. 

8.  Goods  that  cost  me  $820,  I  sell  for  $416.  How  should  I 
mark  goods  that  cost  me  $40  to  make  a  proportional  profit  in 
selling  ? 

Solution:    (1)  Selling  price  of  $40  worth  of  goods  =  $() ?    (Question.) 

(2)  Selling  price  of  $320  worth  of  goods  =  $416.     ( Basis. ) 

(3)  40:320::(  ):416?    (Proportion.) 

(4)  '-^^  =  52. 

320 

.*.  the  goods  should  be  marked  to  sell  for  $52. 


142  ADVANCED   ARITHMETIC. 

9.  I  sell  40  pounds  of  sugar,  16  ounces  to  the  pound,  for  $8.20. 
For  how  much  should  I  sell  40  pounds,  15  ounces  to  the  pound  ? 

Solution :    (1)  Value  of  40  lb.,  15  oz.  to  the  lb.  =  $(  )  ?    (Question.) 

(2)  Value  of  40  lb. ,  16  oz.  to  the  lb.  =  $3.20.    ( Basis.) 

(3)  40  X 15 :  40  X 16 : :  (  ) :  3.2  ?    ( Proportion.) 

,,,  40x15x3.2    „ 

(4)  =3. 

^  '       40x16 

.'.  the  required  value  is  $8. 

Note. — When  the  same  factor  occurs  in  each  part  of  the  problem,  it 
has  no  effect  upon  the  result.  If  the  40's  were  omitted  in  the  above 
problem,  would  that  make  any  change  in  the  result  ? 

10.  A  grocer  sells  a  quantity  of  sugar  by  a  false  weight  of  15 
ounces  to  the  pound  for  $8.20.  How  much  does  he  gain  by  the 
cheat  ? 

Solution :    (1)  Cost  of  *he  number  of  lb.  of  15  oz.  each  =  $(  )  ?    (Question.) 

(2)  Cost  of  same  number  of  lb.  of  16  oz.  each  =  $3.20.    ( Basis.) 

(3)  15 :  16 : !  (  ) :  3.2  ?    ( Proportion.) 

(4)  '^^=3. 

.'.  the  true  selling  price  is  $3,  and  his  gain  by  cheating  is  20^. 

Note. — The  thief  gets  pay  for  true  weight (16  oz.  to  the  lb.).  If  he  had 
delivered  full  weight,  the  sales  would  have  been  worth  $3.20 ;  what  is 
the  amount  worth  which  he  does  deliver  ? 

11.  A  lawyer  who  collects  for  5%,  gets  $84.60  for  collecting  a 
debt.     Find  the  amount  of  the  debt. 

Solution:    (1)  100^  of  debt  =  $(  )?    (Question.) 

(2)  5%  of  debt  =  $34.50.    ( Basis. ) 

(3)  100:5::  (): 34.5?    (Proportion.) 

(4)  ^^?^=m. 

5 

.*.  the  amount  of  the  debt  is  $690. 

12.  A's  property  is  assessed  at  $8800.  What  is  his  tax  at 
96/  on  the  $100  ? 


ADVANCED    ARITHMETIC.  143 

Solution:    (1)  Tax  on  $3800  =  $(  )  ?    (Question.) 

(2)  Tax  on  $100  =  $.96.    (Basis.) 

(3)  3800 :  100: :  (  ) :  .96  ?    (  Proportion.) 

(4)  ??^-^^  =  36.48. 
■       100 

.'.  A's  tax  is  $36.48. 

13.  B  owns  $2500  of  the  capital  stock  of  a  $200000  stock 
company.  The  company  has  $14000  for  distribution  among  its 
stockholders.     How  much  does  B  get  ? 

Solution:    (1)  The  dividend  on  $2500  =  $(  )  ?    (Question.) 

(2)  The  dividend  on  $200000  =  $14000.    (Basis.) 

(3)  2500 :  200000 :  t  (  ) :  14000  ?    (  Proportion. ) 

,,,  2500x14000    ,„^ 
(4) =  170. 

^  '       200000 

.-.  B's  dividend  is  $175. 

14..  A  certain  body  of  soldiers,  standing  125  in  rank  and  60 
in  file,  change  the  file  to  75 :  what  is  the  change  in  rank  ? 

Solution : 

(1)  Length  of  a  certain  body  of  men  75  men  w.  =(  )  men  ?    (Question.) 

(2)  Length  of  same  body  of  men  60  men  w.  =  125  men.    ( Basis.) 
No.    No.  ^2g  ^      Proportion .) 

^  '   75      60  ^ 

NO.X125X60 
^  '       75xNo. 
.'.  the  rank  is  changed  to  100  men. 

Note. — If  there  were  two  different  numbers  of  men,  the  amount  of 
each  would  have  to  be  expressed;  but  as  it  is,  the  same  '^No."  (what- 
ever its  size)  falls  both  above  and  below,  and  does  not  affect  the  result. 

15.  If  800  reams  of  paper  are  consumed  in  printing  88400 
volumes  of  a  160-page  book,  octavo  size,  how  many  reams  of 
paper  will  be  required  in  printing  27000  volumes  of  a  820-page 
book,  duodecimo  size  ? 

Note. — In  making  a  book  octavo  size  each  sheet  is  folded  into  8  leaves, 
making  16  pages ;  in  a  duodecimo  size  each  sheet  is  folded  into  12  leaves, 
making  24  pages. 


144  ADVANCED    ARITHMETIC, 

Solution : 

(1)  Amt.  of  paper  reqd.  for  27000  vol.  of  320  pp.  each  =  (  )  reams  of  24 

pp.  to  the  sheet  ?    (Question.) 

(2)  Amt.  of  paper  reqd.  for  38400  vol.  of  160  pp.  each  =  800  reams  of  16 

pp.  to  the  sheet.    ( Basis.) 

(3)  27000  X  320 :  38400  x  160 : :  (  )  x  24 :  800  x  16  ?     ( Proportion .) 

,^.  27000x320x800x16    ->^      . 

(4) =  7oU. 

'       38400x160x24 

.*.  750  reams  will  be  required. 

16.  What  rate  is  required  for  $600  to  produce  $192  interest  in 
4  years  ? 

Solution :    (1)  Int.  on  $600  for  4  yr.  at  (  )^  =  $192  ?    (Question.) 

(2)  Int.  on  $1  for  1  yr.  at  1^  =  $.01.    (Basis.) 

(3)  600x4x(  ):  1x1x1::  192:. 01?    (Proportion.) 
1x1x1x192^19200^ 

600X4X.01  ~  2400       * 
.•.  the  reqd.  rate  is  8^. 

17.  What  principal  will  be  required  to  bear  $225  at  9%  in 
3J  years  ? 

Solution  :    (1)  Int.  on  $(   )  for  3^  yr.  at  9^  =  $225  ?    (Question.) 

(2)  Int.  on  $1  for  1  yr.  at  1^  =  $.01,    ( Basis.) 

(3)  (  )x3ix9: 1x1x1:: 225 :.01?    (Proportion.) 

(4)  lxlxlx225^y,Q^ 

3^x9x.01 
.*.  the  required  principal  is  $750. 

18.  If  8  men  mow  36  acres  in  9  days  of  9  hours  each,  how 
many  men  can  mow  48  acres  in  12  days  of  12  hours  each  ? 

Solution : 

(!)  Amt.  mowed  by  (  )men  in  12  da.  of  12  hr.  each  =  48  A.  ?    (Question.) 

(2)  Amt.  mowed  by  8  men  in  9  da.  of  9  hr.  each  =  36  A.     (Basis.) 

(3)  (  )xl2xl2:8x9x9::48:36?    (Proportion.) 
,  .  8x9x9x48_ 

12x12x36  ~ 
.'.  the  required  answer  is  6  men. 


ADVANCED   ARITHMETIC.  145 

19.  A  garrison  of  men  have  food  to  last  9  months,  giving 
each  man  1  pound  2  ounces  per  day.  What  should  be  the  daily 
allowance  to  make  the  same  food  last  1  year  8  months  ? 

Note.— (1)  1  lb.  2  oz.'=L8  oz.  ;  1  yr.  8  mo.  =  20  mo. 

Solution  : 

W  I  in  20^Z'  iTn  ol  Srh'^pTr  da.  |  ^  «-*-"  ">»  '   (Q"-«-) 

(2)  1  in  /r^^tTo'i'i^^lTr  da.  |  —  -"unt.    (Basis.) 

(3)  20  X  (  ) :  9  X 18 : :  Amt. :  Amt.  ?    (Proportion.) 
...  9 X 18 X amt.     „  , 

20  X  amt. 
.*.  the  required  answer  is  8^^^  oz. 

SO.  If  50  men  in  10  days  of  9  hours  each,  build  a  wall  250 
yards  long,  8  feet  high,  3  feet  thick,  how  many  hours  per  day 
must  75  men  work  to  build  a  wall  650  yards  long,  9  feet  high, 
4  feet  thick,  in  80  days  ? 

{Amt.  built  by  75  men)       f  *  ,„oii  acn  ^;i   i 
working  30  da.  of         [  =  \  ^.^  ,     f  f/^i^V '  /n„^cf  ^^ 
()hr.each  J      |9  ft.  h.,  4  ft.  th?    (Question.; 

{Amt.  built  by  50  men ")       f  »  „.„n  oka  ^a   i 
^Z''.7.r-'        hl^a'^^TA!- (Basis.) 

(3)  75x30x(  ):50xlOx9::650x9x4:250x8x3?     (Proportion.) 
.  .  50xlOx9x650x9x4_ 

75x30x250x8x3    ~   " 
.•.  the  required  answer  is  7.8  hr. 

EXERCISE  LVII. 

1.  Find  the  cost  of  150  bushels  of  apples  at  $.65  per  bushel. 

2.  If  12  tons  of  hay  can  be  bought  for  $36,  how  much  must 
be  paid  for  50  tons  ? 

3.  A  farmer  sows  6  bushels  of  grain  on  4f  acres.  At  that 
rate,  how  many  bushels  will  be  needed  to  sow  3.6  acres  ? 

4..  If  it  requires  42  yd.  of  carpet  |  yd.  wide  to  cover  a  floor, 
how  many  yards  of  carpet  1  y-d.  wide  will  be  needed  ? 


146  ADVANCED   ARITHMETIC. 

5.  In  preparing  a  certain  mixture,  12  gills  of  water  were 
mixed  with  7  gills  of  other  ingredients.  How  many  gills  of 
other  ingredients  should  be  mixed  with  42  gills  of  water  ? 

6.  15%  of  a  debt  is  $16.50.     Find  80%  of  it. 

7.  A  grocer  uses  false  weight  whereby  Ib^  oz.  is  sold  for  a 
pound.  What  is  the  true  value  of  groceries  which  he  sells  for 
$320? 

8.  If  20  yd.  of  cloth  1.5  yd.  wide  sell  for  $15,  how  wide 
must  51  yd.  of  the  same  kind  of  cloth  be  to  sell  for  $42.50  ? 

9.  What  are  a  servant's  wages  for  3  weeks  5  days,  at  $1.75 
per  week  ? 

10.  A  body  of  soldiers  are  42  in  rank  and  24  in  file.  If  they 
were  changed  to  36  in  rank,  how  many  in  file  would  they  be  ? 

11.  An  equatorial  degree  is  865000  feet.  How  many  feet  in 
75°  24'',  measured  on  the  equator  ? 

12.  If  a  troy  pound  of  English  standard  silver  is  worth  £3^, 
what  is  1  lb.  av.  worth  ? 

IS.  If  coffee  which  cost  $225  is  now  worth  $318.75,  what  was 
the  cost  of  coffee  now  worth  $1285.20  ? 

IJ^.  A  bin  7  ft.  long,  2^  ft.  wide,  2  ft.  high,  contains  28  bu. 
of  corn.  How  deep  must  one  be  which  is  18  ft.  long,  4f  ft. 
wide,  to  contain  120  bushels  ? 

15.  If  150000  bricks  are  used  in  building  a  house  whose  walls 
average  1^  ft.  thick,  80  ft.  high,  and  all  together  are  216  ft.  long,/ 
how  many  bricks  will  be  required  to  build  a  wall  2  ft.  thick,  24 
ft.  high,  and  324  ft.  long  ? 

16.  It  requires  800  reams  of  paper  to  print  28800  volumes  of 
a  duodecimo  book  of  320  pages  to  the  volume.  How  many 
reams  of  paper  will  be  required  to  print  24000  volumes  of  an 
octavo  book  of  520  pages  to  the  volume  ? 

17.  The  premium  on  a  draft  for  $4320  is  $21.60:  how  large  a 
draft  can  I  buy  at  a  premium  of  $32.10  ? 

18.  The  rate  of  premium  as  in  No.  17.  What  will  be  the 
premium  on  a  draft  for  $1284.20  ? 


ADVANCED    ARITHMETIC.  147 

19.  A  property  worth  $2540  rents  for  $139.70:  at  that  rate, 
what  should  property  valued  at  $4000  rent  for  ? 

20.  At  the  rate  in  No.  19,  what  is  the  value  of  property  which 

rentsfor  $192.70?  ; 

21.  What  is  the  tax  on  $3450  worth  of  property  at  $1.60  on 
the  $100  ? 

22.  If  A  charges  5/  on  the  dollar  for  collecting  a  note  and 
receives  as  his  commission  $24.50,  what  is  the  face  of  the  note  ? 

23.  A  man  failing  is  able  to  pay  only  $65  on  the  $100  of  his 
debts.     What  does  E  get,  if  this  man  owes  him  $824  ? 

24..  A  stock  company  whose  capital  is  $4256000  is  ready  to 
distribute  $148960  among  its  stockholders.  Smith  gets  a  divi- 
dend of  $157.50:  how  much  stock  has  he  ? 

25.  If  16  men  build  18  rods  of  fence  in  12  days,  how  many 
men  can  build  72  rods  in  8  days  ? 

26.  If  $100  gain  $8  in  12  months,  how  many  dollars  will  gain 
$144  in  4  months  ? 

27.  If  6  men  spend  $150  in  8  months,  how  much  will  15  men 
spend  in  20  months  ? 

28.  If  180  men  in  6  days  of  10  hours  each  dig  a  trench  200 
yd.  long,  3  yd.  wide,  2  yd.  deep,  in  how  many  days  of  8  hr.  each 
can  100  men  dig  a  trench  180  yd.  long,  4  yd.  wide,  3  yd.  deep  ? 

29.  If  a  regiment  of  840  men  has  food  for  60  days,  how  many 
days  should  the  same  food  last  a  garrison  of  1260  men  ? 

30.  If  a  pipe  discharging  4  gallons  per  minute  fill  a  cistern 
in  2  hours,  in  how  many  minutes  may  the  cistern  be  filled  by 
a  pipe  discharging  9  gallons  per  minute  ? 

31.  If  the  use  of  $1500  for  3  yr.  8  mo.  25  da.  is  worth  $336.25, 
what  is  the  use  of  $100  for  1  yr.  worth  ? 

32.  If  4  horses  draw  a  car  9  miles  per  hour,  how  many  miles 
per  hour  will  an  engine  of  150  horse-power  draw  a  train  of  12 
such  cars,  adding  its  own  weight,  3  cars  ? 


PART  II. 

I.    STUDY  OF  NUMBERS. 
A.     DENOMINATE  NUMBERS. 

82.  Definitions. — A  denominate  number  repre- 
sents quantity  as  composed  of  units  of  a  particular  denomina- 
tion or  kind.     As, 

50  bushels,         30  minutes,         8  pounds. 

Most  magnitudes  may  be  expressed  in  units  of  two  or  more 
sizes.  Thus,  time  may  be  expressed  in  seconds,  minutes,  hours, 
etc. 

The  process  of  determining  the  number  of  units  of  one  size 
in  a  number  expressed  in  units  of  another  size,  is  called  re- 
duction.    As, 

(1)  To  find  how  many  quarts  in  10  gallons;  or, 

(2)  To  find  how  many  gallons  in  40  quarts. 

When  the  change  is  from  a  larger  (higher)  to  a  smaller 
(lower)  unit,  as  in  (1),  the  reduction  is  descending;  when 
from  a  smaller  to  a  larger  unit,  as  in  (2),  the  reduction  is  as- 
cending. 

In  order  to  change  or  reduce  a  number  from  units  of  one  size 
to  units  of  another,  the  relative  size  of  these  units  must  be 
known.  The  equation  which  states  this  relation  is  the  basis  of 
solution.  These  equations  are  given  in  the  tables.  The  plan 
and  process  of  solution  are  the  same  as  learned  in  Part  I. 

(148) 


ADVANCED    ARITHMETIC.  149 

There  are  two  systems  of  measurement  in  use  in  the  United 
States,  The  English  System  and  The  French  System. 

The  English  System,  originated  by  the  English,  has  come 
into  common  use  among  the  people  of  the  United  States  for 
all  ordinary  measurements  and  computations ;  except  the  Eng- 
lish units  of  value,  which  are  not  used  in  the  United  States. 

The  French  System,  originated  by  the  French,  is  used  in  the 
United  States  principally  by  scientists  in  making  scientific 
measurements  and  computations. 

I.     ENQIilSH  SYSTEM. 

83.  Linear  Measures. — A  Liine  is  that  which  has 
only  one  dimension,  length.  The  linear  measures,  or  units,  are 
used  in  measuring  heights,  lengths,  widths,  distances,  etc. 
There  are  two  tables,  the  Conimon  Ijinear  Table  and 
the  Surveyors'  Linear  Table. 

COMMON  lilNEAR  TABIiB. 

1  yard  (yd.)  =3  feet  (ft.). 
1  ft.  =  12  inclies  (in.). 

SURVEYORS'  lilNEAR  TABLE. 

1  mile  (mi.)  =80  cliains  (ch.). 

(  4  rods  (rd.),  or 
1  cliain=  •<  66  ft.,  or 

(  100  links  (1.). 

1  rod=  I  ^^'^'  ^t"  ^^ 
I  25  I. 

Note  1. — A  link  is  7.92  inches  long.  It  is  seldom,  if  ever,  necessary  in 
practice  to  reduce  links  to  inches  or  inches  to  links. 

Note  2. — In  ordinary  land-surveying,  surveyors  use  a  chain  100  links, 
or  66  feet,  long, — sometimes  called  Gunter's  Chain.  Civil  engineers  gen- 
erally use  a  ''steel  tape,"  100  feet  long. 


150  ADVANCED   ARITHMETIC. 

EXAMPIiES. 

1.  Reduce  5  yd.  2  ft.  9  in.  to  inches. 

Plan.— (1)  Reduce  the  yards  to  feet,  and  add  the  2  ft.  (2)  Reduce  the 
feet  to  inches,  and  add  the  9  in.  Each  step  contains  one  problem  and 
one  addition. 

Solution  :     (1)  5  yd.  =  (  )  ft.?    (Question.) 
(2)  1yd.  =  3  ft.    (Basis.) 
5x(2)  =  (3)  5  yd.  =  15  ft. 

(4)  15  ft. +2  ft.  =  17  ft. 


(5)  17ft.  =  (  )in.?    (Question.) 

(6)  1ft.  =  12  in.     (Basis.) 
17x(6)  =  (7)  17  ft.  =204  in. 

(8)  204  in.  +  9  in .  =  213  in . ,  answer. 

2.  Reduce  2  mi.  20  ch.  74  1.  to  links. 

KoTE. — If,  in  the  judgment  of  the  teacher,  the  solution  is  sufficiently 
clear  to  the  pupil  without  stating  the  ^'Question"  each  time,  the  '^Ques- 
tions' may  be  omitted.    Thus, 

Solution:    (1)  1  mi.  =  80  ch.     (Basis.) 
2x(l)  =  (2)  2mi.  =  160ch. 

(3)  160ch.+20ch.  =180ch. 


(4)  lch.  =  1001.    (Basis.) 

180 X (4)  =  (5)  180  ch.  =  180001. 

(6)  18000  1.  +  74  1.  =  18074  1.,  answer. 

3.  Prepare  the  surveyors'  linear  table  for  use  in  reduction 
ascending. 

Note. — In  reduction  descending,  the  order  of  the  process  is  from 
higher  to  lower  units,  and  the  larger  unit  in  each  equation  is  on  the 
left.  In  reduction  ascending,  this  order  is  reversed,  and  the  smaller 
unit  is  placed  on  the  left.    Thus, 

1    '  1    1  25  1 

1  ^"•= 7:921- =r98i- 

1  1  _  i  xiiy  ch. 


ADVANCED    ARITHMETIC.  151 

lrd.  =  ich. 
-   1  ch.  =  ^  mi. 

If..  Reduce  24848  1.  to  higher  units. 

Solution:    (1)  1 1.  =T^Ty  ch.     (Basis.) 

24348  X  (1)  =  (2)  24348  1.  =  HU-  ch.  =  243  ch. +48  1. 

(3)  1  ch.  =^  mi.     (Basis.) 
243 X (3)  =  (4)  243ch.  =  3mi.  +  3ch. 

.-.  243481.  =3  mi.  3  ch.  48  1. 

Another  Solution  :    (1)  11.  =  ^V  rd.    ( Basis.) 
24348  X (1)  =  (2)  24348  1.  =  973  rd.  +  23  1. 

(3)  lrd.  =  ich.    (Basis.) 
973  X (3)  =  (4)  973  rd.  =  243  ch.  + 1  rd. 

(5)lch.  =  ^mi.    (Basis.) 
243x(5)  =  (6)  243ch.  =  3mi.  +  3ch. 

.-.  24348  1.  =3  mi.  3  ch.  1  rd.  23  1. 

Note. — In  the  first  solution,  equation  (2),  ^U-  ch.  is  simplified  by  di- 
viding the  24348  by  100.  The  quotient  is  243  ch. ,  the  remainder  is  yVif  ch. 
But  1^  ch.  is  1 1.,  therefore  iVir  ch.  =48  1.  Thus,  the  quotient  represents 
the  number  of  chains,  and  the  remainder  the  number  of  links. 

5.  Reduce  444  rd.  to  miles. 

Solution:    (l)lrd.=ich.    (Basis.) 
444  X  (1)  =  (2)  444  rd.  =  111  ch. 

(3)lch.  =  ^mi.    (Basis.) 
Ill  X (3)  =  (4)  111  ch.  =  l|^  mi.,  or,  1.3875  mi.,  answer. 

Note. — In  reducing  to  ^^ higher  units,"  the  remainders  are  left  as  inte- 
gers, and  the  quotient  only  is  reduced  higher.  In  reducing  to  a  particu- 
lar denomination,  as  in  No.  5,  the  exact  quotient  is  obtained  each  time  and 
all  carried  to  the  required  denomination.  In  reducing  fractions  to 
'Hower  units,'''  when  the  result  is  a  mixed  number,  leave  the  integral 
part,  and  reduce  only  the  fraction  lower. 


152  ADVANCED    ARITHMETIC. 


6.  Reduce  f  yd.  i  ft.  to  inches. 

Solution:    (1)1  yd.  =3  ft.    (Basis.) 
|of(l)  =  (2)  I  yd.  =2  ft. 

(3)  2ft.  +  ift.  =  2ift. 


(4)  1ft.  =  12  in.     (Basis.) 
2i  X  (4)  =  (5)  24  ft.  =  27  in.,  answer. 

7.  Reduce  ^  it.  to  the  fraction  of  a  mile. 

Solution:    (l)lft.  =^ch,    (Basis.) 
^of(l)  =  (2)  x\ft.  =  TTrWch. 


(3)  1  ch.  =  gi(5  mi.     (Basis.) 
o^ffe  of  (3)  =  (4)  xTrVe  ch.  =  lefgr  mi.,  answer. 

8.  Reduce  .32  rd.  to  the  decimal  of  a  mile. 

Solution :    (1)  1  rd.  =  i  ch.  =  .25  ch. 
.32of  (1)  =  (2)  .32rd.  =  .08ch. 


(3)  1  ch.  =^xr  mi.  =  .0125  mi. 
.08of  (3)  =  (4)  .08  ch.  =  .001  mi.,  answer. 

Note. — When  a  decimal  result  is  required,  it  may  be  obtained  as 
above  by  expressing  all  fractions  as  decimals ;  or  the  solution  may  be 
expressed  in  common  fractions  and  the  result  only  reduced  to  a  decimal. 

9.  Find  in  integral  units  the  value  of  |  mi.  minus  ^  rd. 

Solution:    (1)  1  mi.  =  80  ch.     (Basis.) 
f  of  (1)  =  (2)  f  mi.  =  17|ch. 


(3)  lch.=4rd.    (Basis.) 
iot  (3)  =  (4)  |ch.  =  -2/rd. 

(5)  -%^  rd.  -  j\  rd.  =  %V  rd.  - U  rd.  =  W  rd.  =  2^  rd. 


(6)  1  rd.  =  161  ft.  =-3/  ft.    ( Basis.) 
if  of  (6)  =  (7)  i|rd.  =  ex-¥-ft.  =  12Ht. 


(8)  1ft.  =  12  in.    (Basis.) 
iof  (8)  =  (9)  i  ft.  =  2  in. 

.-.  f  mi.  -  /x  rd.  =17  ch.  2  rd.  12  ft.  2  in. 


ADVANCED   ARITHMETIC.  153 

10.  Reduce  ^  ft.  |  rd.  |  ch.  to  the  fraction  of  a  mile. 

Solution :    (1)  1  ft.  =  i^  rd.    ( Basis.) 
(2)  Jft.  =  s^jrd. 
,<a)  ^s  rd.  +  l  rd.  =  xh-  v^.  +  ^i^  rd.  =  I§|  rd. 


(4)  lrd.=ich.    (Basis.) 

(5)  ieird.  =  ^2f  ch. 

(6)  m  ch.  +  i  ch.  =  m  ch.  +  IM  ch.  =  til  ch. 


(7)  1  ch.  =  ^mi.    (Basis.) 

(8)  tM  ch.  =^i^  mi.,  answer. 

Note. — Notice  that  number  10  may  be  stated  in  the  form  of  an  exam- 
ple of  addition.    Thus,  ^  f t.  + 1  yd.  + 1  ch.  =  (  )  mi. ? 


EXERCISE  LVIII. 

1.  Reduce  1  mi.  to  rods. 

2.  Reduce  1  mi.  to  feet. 

8.  Reduce  7  mi.  240  rd.  to  inches. 
J/..  Reduce  8  mi.  79  ch.  to  links. 

5.  How  many  yards  in  1  mile  ? 

Plan.— (1)  Reduce  1  mi.  to  ft.,  (2)  reduce  the  ft.  to  yd, 

6.  Reduce  W  mi  to  integral  units. 

7.  Reduce  -J^J  mi.  /^  ch.  f  rd.  i  ft.  to  inches. 

8.  Reduce  J|  mi.  y\  ch.  f  rd.  to  integral  units. 

9.  Reduce  .35  mi.  7.3  ch.  to  inches. 

10.  Reduce  .37  mi.  5.7  ch.  to  links. 

11.  Reduce  340000  1.  to  miles. 

12.  Reduce  248000  in.  to  higher  units. 
18.  Reduce  12480  ft.  to  higher  units. 

IJ^.  Reduce  25  ft.  to  the  fraction  of  a  chain. 
15.  Reduce  83  ft.  to  the  fraction  of  a  mile. 


154 


ADVANCED   ARITHMETIC. 


16.  Reduce  5.5  ft.  to  the  decimal  of  a  mile. 

17.  Reduce  y'^  1.  to  the  fraction  of  a  mile. 

18.  Reduce  1  1.  1  ft.  1  rd.  1  ch.  to  the  fraction  of  a  mile. 

19.  y\ch.-y\rd.  =  (   )  mi.? 

Plan. — (1)  Reduce  the  ch.  to  rd,  and  subtract;  (2)  reduce  the  result 
to  mi. 


20.  ^  mi.+-^  ch.+f  ft.=  (   )  in.? 

21.  fmi.— iVch.+|rd.-2ft.  =  (   )  in.? 

22.  5  ch.-3.5  rd.+4i  ft. -11  in.  =  (  )  mi.? 

84.  Surface  Measures.— A  Surface  is  that  which 
has  only  two  dimensions,  length  and  ivklth.  The  number  of 
square  units  in  a  surface  is  its  Area.  A  Square  Unit  is 
the  amount  of  a  surface  1  unit  long  and  1  unit  wide. 


Square  Unit. 

There  are  two  tables  of  surface  measures,  corresponding  to 
the  two  tables  of  linear  measures.  The  one  is  the  Comraon 
Square  Measures,  used  to  measure  the  area  of  such  sur- 
faces as  floors,  ceilings,  carpets,  cloth,  and  the  like. 

TABIiR 

1  sq.  yd.  =  9  sq.  ft. 
1  sq.  ft.  =  144  sq.  in. 


The  other  table  is  the  Surveyors'  Square  Measures, 
used  in  measuring  the  area  of  land. 


ADVANCED   ARITHMETIC.  155 


TABIiE. 


1  to^wnship  (Tp.)=36  sq.  mi.,  or  Sections  (Sec.) 
1  sq.  ini.=:64Q  acres  (A.). 


1  A.=  I  160  sq.  rd. 

(  10  sq.  en. 
1  sq.  ch.  =  16  sq.  rd. 

1  sq.  rd. =2721/4  sq.  ft. 


EXERCISE  LIX. 

1.  How  many  acres  in  a  section  ?  ^ 

2.  How  many  acres  in  a  township  ? 

3.  How  many  square  rods  in  a  square  mile  ? 

4.  Reduce  12  sq.  yd.  7  sq.  ft.  to  square  inches. 

5.  Reduce  25  sq.  yd.  5  sq.  ft.  to  square  inches. 

6.  Fill  the  following  blanks,  preparing  the  table  for  work  in 
reduction  ascending : 

1  sq.  ft.  =TW¥  sq.  rd. 

(   )  sq.  ch.? 


^^•^^-  U   )A.? 
Isq.  ch.  =  (   )  A.? 
1  A.  =  (   )  Sec? 
1  Sec.  =  (   )Tp.? 

Also, 

1  sq.  in.=  (    )  sq.  ffc.? 
1  sq.  ft.=  (   )  sq.  yd.? 

7.  Reduce  2592  sq.  in.  to  sq.  yd. 

8.  Reduce  32000  A.  to  sections. 

9.  How  many  townships  contain  648  sq.  mi.? 

10.  How  many  square  miles  in  231200  sq.  rd.? 

11.  How  many  sq.  ch.  in  2  Sec? 

12.  Reduce  7  sq.  mi.  120  A.  150  sq.  rd.  to  square  rods. 


156 


ADVANCED    ARITHMETIC. 


13.  Reduce  202480  sq.  rd.  to  higher  units. 

IJ^.  Reduce  |  A.  to  lower  units. 

15.  Reduce  .625  sq.  rd.  to  lower  units. 

16.  Reduce  -j\  sq.  yd.  to  sq.  inches. 

17.  Reduce  10  sq.  ft.  to  the  fraction  of  a  sq.  yd. 

18.  Reduce  y\^^  sq.  in.  to  the  decimal  of  a  sq.  yd. 

19.  I  sq.  mi.  +  .5  A. +  120  sq.  rd.  =  (   )  sq.  rd.? 

20.  12.5  sq.  ch.+6i  sq.  rd.  =  (   )  sq.  mi.? 

21.  .6sq.  yd.  =  (   )  A.? 

Plan. — (1)  Reduce  sq.  yd.  to  sq.  ft. ;  (2)  reduce  sq.  ft.  to  sq.  rd. ;  (3)  re- 
duce sq.  rd.  to  A. 


?.  -,3^A.-|sq 

7 


rd.+240sq.  ft. 


(   )  sq.  in. 

Tir  A.-l  sq.  rd.+240  sq.  ft.  =  (   )  Sec? 
^^.  5sq.  mi.-20.4  A.+T^jsq.  ch.  =  (   )  Tp.? 


85.  Solid  Measures. — A  Solid  is  that  which  has 
three  dimensions,  length,  width,  and  thickness.  The  number  of 
cubic  units  in  a  solid  is  its  Volume.  A  Cubic  Unit  is 
the  amount  of  a  solid  1  unit  long,  1  unit  wide  and  1  unit 
thick. 


Cubic  Unit. 

The  Cubic  Measures  are  used  in  measuring  volumes  of  solids, 
and  the  capacities  of  bins,  tanks,  and  the  like. 


1  cu.  yd.  =  27  cu.  ft. 
1  cu.  ft.  =  1728  cu.  in. 


ADVANCED    ARITHMETIC.  157 

EXERCISE  LX. 

1.  Reduce  5  cu.  yd.  15  cu.  ft.  to  cu.  in. 

2.  A  cord  of  wood  is  8  ft.  long,  4  ft.  wide,  and  4  ft.  high. 
How  many  cubic  feet' does  it  contain  ? 

Basis  :  Vol.  of  a  solid  1  ft.  1.,  1  ft.  w. ,  1  ft.  h.  =  1  cu.  ft. 

3.  How  many  cubic  inches  in  a  cord  ? 
^.  Reduce  5  cords  (C.)  to  cubic  ft. 

5.  288  cu.  in.  is  what  part  of  a  cubic  yd.? 

Note.— Prepare  the  table  for  reduction  ascending  before  solving  No.  5 . 

6.  12  cu.  ft.  6  cu.  in.  is  what  part  of  1  C? 

7.  Reduce  ^  cu.  ft.  to  cu.  yd. 

8.  I  cu.  yd.+24|  cu.  ft.  =  (   )  cu.  in.? 

9.  A  perch  of  masonry  is  16^  ft.  long,  1^  ft.  thick,  and  1  ft. 
high.     How  many  cubic  feet  does  it  contain  ? 

10.  If  cu.  ft.  — A  cu.  in.  =  (   )  cu.  yd.? 


86.  Measures  of  Capacity. —  The  extent  of  room  or 
space  within  a  vessel  is  called  its  Cax^acity.  Measures  of 
capacity  are  classed  as  Dry  Measures  and  Liquid  Measures.  The 
dry  measures  are  used  to  measure  quantities  of  grain,  fruits, 
seeds,  and  the  like. 

TABLE. 

1  bushel  (bu.)=4  pecks  (pk.). 
1  x>k.  =  8  quarts  (qt.). 
1  qt.  =  2  pints  (pt.). 

Note. — 1  bushel  contains  2150.4+  cu.  in.  Fruits,  vegetables,  seeds 
and  the  like  are  bought  and  sold  by  the  bushel ;  but  the  amount  of  the 
bushel  is  often  expressed  by  weight  rather  than  size.    .(See  p.  161.) 


158  ADVANCED   ARITHMETIC. 

EXERCISE  LXI. 

1.  Reduce  20  bu.  to  pints. 

2.  Reduce  2  bu.  2  pk.  to  quarts. 

3.  Reduce  7  bu.  8  pk.  7  qt.  to  pints. 

4.  Reduce  9  bu.  1  pk.  4  qt.  1  pt.  to  pints. 

5.  Reduce  768  pt.  to  bushels. 

6.  Reduce  679  pt.  to  higher  units. 

7.  Reduce  f  bu.  to  pints. 

8.  Reduce  -J^  bu.  to  lower  units. 

9.  |bu.+|pk.+|qt.+ipt.  =  (  )pt.? 

10.  ibu.+}pk.+|qt.+ipt.  =  (  )bu.? 

11.  5bu.-ipk.-Hqt.-Hpt.  =  (   )pt.? 

12.  ibu.-ipk.-|.375qt.-lipt.  =  (   )bu.? 

13.  Find  the  value  of  4  bu.  8  pk.  of  apples  at  20/  per  peck. 
IJ/..  A  grocer  buys  ^  bu.  of  cherries ;   he  then  sells  1  bu.  8 

pk.  of  them;  again,  he  buys  1  bu.  1  pk.  8  qt„  of  cherries; 
finally  he  sells  all  the  cherries  he  has  at  6/  per  quart.  Find 
the  value  of  the  last  sale. 

There  are  two  tables  of  Liquid  Measures  in  use,  (1)  the  Com- 
mon Liquid  Measures,  used  in  measuring  such  liquids  as  water 
and  milk,  and  (2)  the  Apothecaries^  Liquid  Measures^  used  by 
apothecaries  in  measuring  liquid  medicines : 

COMMON  TABLE. 

1  liog"slieacl  (]i]id.)=2  barrels  (bbl.)o 

1  bbl.  =  311/2  gallons  (g-al.)o 

1  gal. =4  qt. 

1  qt.  =  2  pt. 

1  pt.  =  4  gills  (gi.). 

KoTE. — 1  gallon  contains  231  cu.  in.  The  dry  measure  quart  contains 
9.45  cu.  in.  more  than  the  liquid  quaint. 


ADVANCED    ARITHMETIC.  159 

APOTHECAKTES'  TABLE. 

1  gallon  (*Coiig.)  =  8  pints  (*0.) 
1  O.  =  10  fluid  ounces  (fB)o 
1  fS  =  8  fluid  drams  (f5). 
1  f5  =  60  minims  (m.). 

EXERCISE  LXII, 

1,  Reduce  1  hhd.  to  gills. 

2.  Reduce  2  hhd.  1  bbl.  21  gal.  to  quarts. 
S,  Reduce  ^  gal.  -J  qt.  to  gills. 

^.  Reduce  1008  gi.  to  barrels. 

5.  Reduce  1008  pt.  to  higher  units. 

6.  Reduce  5  Cong,  to  minims. 

7.  Reduce  5  O.  2  f  5  3  f  5  to  minims. 

8.  Reduce  1260  m.  to  higher  units. 

9.  A  physician  wishes  to  prepare  a  5%  solution  of  carbolic 
acid.  How  much  water  and  how  much  acid  must  be  used  to 
make  2  fluid  ounces  ? 

Note. — h%  of  the  2f5  is  carbolic  acid,  and  the  rest  is  water. 

10.  What  will  1  bbl.  of  syrup  sell  for,  at  80/  per  gallon? 

87.  Measures  of  Mass. — The  Mass  of  a  body  is  the 
amount  of  matter  it  contains.  The  Weig-lit  is  the  measure 
of  the  attraction  between  that  body  and  the  earth.  Weights  are 
used  to  measure  the  mass  of  a  body. 

Note. — To  be  accurate  in  comparing  the  masses  of  bodies  by  their 
weights,  the  weights  must  be  taken  at  the  same  altitude,  in  the  same 
latitude,  and  under  the  same  conditions.  But  for  ordinary  purposes,  this 
accuracy  is  not  observed. 


*  The  Latin  words  are  Congius  and  Octavua. 


160  ADVANCED   ARITHMETIC. 

There  are  three  tables  of  weights :  (1)  The  table  of  Avoir- 
dupois Weights,  used  in  weighing  all  ordinary  articles, 
such  as  groceries,  meats,  live  stock,  etc. 


1  ton.  (T.)=20  hundred weiglit  (cwt.). 
1  c\v^t.  =  100  i^ounds  (lb,)e 
1  lb.  =  16  ounces  (oz.). 

(2)  The  table  of  Troy  Weights,  used  in  jeivelry  stores ^ 
and  at  mints,  when  the  Government  weighs  gold  and  silver  for 
making  money. 

TABIiE. 

1  pound  (lb.)  =  12  ounces  (oz.). 
1  oz.  =  20  penny^veiglits  (pwt.). 


1  oz.  =  20  penny^veiglits 
1  pwt.  =  24  grains  (gr.). 


(3)  The  table  of  Apothecaries'  Weights,  used  by 
druggists  in  filling  prescriptions. 

TABLE. 

1  pound  (lb.  =  12  ounces  (.^)a 
1  S  =  8  drams  (5)- 
1  5  =  3  scruples  (9) 
1  9=20  grains  (gr,). 

Note. — ^The  grain,  ounce,  and  pound  of  the  troy  weights  are  identical 
with  the  grain,  ounce,  and  pound,  respectively,  of  the  apothecaries' 
weights.  The  troy  or  apothecaries'  pound  contains  5760  gr.,  but  the  av- 
oirdupois pound  contains  7000  gr. 

Grain  and  vegetables  are  often  bought  and  sold  by  weight- 


ADVANCED   ARITHMETIC.  161 

bushels  instead  of  the  measured  bushels.     In  the  majority  of 
States  the  bhshel  is  of  the  following  weight : 

^  '^     .  TABLE. 

1  bu.  of  wheat =60  lb. 

1  bu.  of  corii=56  lb. 

1  bu.  of  oats  =  32  lb. 

1  bu.  of  Irish  potatoes  =  60  lb. 

1  bu.  of  sweet  j)otatoes=55  lb. 

Note. — The  pounds  of  the  above  table  are  avoirdupois  pounds.  These 
equivalents  are  not  the  same  in  all  States,  and  the  teacher  should  teach 
the  pupil  the  table  corrected  for  his  own  State. 

EXERCISE  LXIII. 

1.  Reduce  1  T.  3  cwt.  4  lb.  to  ounces. 

2.  Reduce  7000  oz.  to  higher  units. 

3.  Reduce  1  lb.  45  55  23  17  gr.  to  grains. 

4.  Reduce  6743  gr.  troy  to  higher  units. 

6.  Reduce  6743  gr.  apothecaries'  to  higher  units. 

6.  Reduce  1  lb.  avoirdupois  to  highest  integral  units  of  the 
troy  table. 

7.  Reduce  2  lb.  apothecaries'  to  highest  integral  units  of  the 
avoirdupois  table. 

8.  A  man  sells  9600  lb.  of  Irish  potatoes  at  70/  per  bu.  How 
much  do  the  potatoes  bring  ? 

9.  A  silver  dollar  weighs  17  pwt.  4^  gr.  What  is  the  weight 
of  $80  in  silver  ? 

10.  A  druggist  buys  lactopeptine  at  $8  per  lb.  and  sells  it  at 
12|/  per  5.     What  is  his  per  cent  of  gain  ? 

11.  How  many  bales  of  70  lb.  each  in  14  T.  17  cwt.  50  lb.  of 
hay? 

12.  A  man  sold  3  loads  of  corn  of  2352  lb.  each.  How  many 
bushels  did  he  sell  ? 


162  ADVANCED   ARITHMETIC. 

88 •  Measures  of  Time. — These  measures  are  used  in 
giving  the  time  of  day,  in  Longitude  and  Time  problems,  and 
in  computing  the  interest  on  notes  and  bills. 


1  year  (jt.)  =  12  montlis  (ino.)« 

1  mo.  =  30  days  (da.), 

1  da. =24  hours  (lir.). 

1  lir.  =  60  minutes  (min.). 

1  min.  =  60  seconds  (sec). 

Calendar  Months: 

January,  81  days.  July,  31  days. 

♦February,  28  days.  August,  81  days. 

March,  81  days.  September,  80  days. 

April,  80  days.  October,  81  days. 

May,  81  days.  November,  80  days. 

June,  80  days.  December,  81  days. 

Note. — The  above  data  give  three  different  lengths  to  the  year.  (1) 
From  the  table,  1  yr.  =  12  mo.  =360  da.  This  may  be  called  the  common 
interest  year.  The  custom  in  the  majority  of  the  States  is  to  use  360  da. 
for  a  year.  (2)  By  counting  the  number  of  days  of  the  calendar  months, 
the  length  of  the  year  is  found  to  be  365  days.  This  is  called  the  com- 
mon year.  (3)  Leap  Year. —  The  average  length  of  a  year  (a  complete 
revolution  of  the  earth  around  the  sun),  expressed  in  mean  solar  time 
(sun  time)  is  365  days,  5  hours,  48  minutes,  47.8  seconds.  The  Julian 
Calendar  called  the  year  365^  days.  For  ordinary  purposes  it  is 
best  to  use  whole  days;  so  the  J^  day  each  year  amounted  in  4 
years  to  1  day.  In  this  way,  the  Julian  Calendar  had  3  common  years 
of  365  days  each,  then  1  leap  year  of  366  days.  The  extra  day  was 
added  to  February.  This  calendar,  by  counting  3653^  days  for  a 
year,  used  11  minutes  12.2  seconds  too  much.  The  Gregorian  Cal- 
endar corrected  this  error  by  omitting  the  leap  year  on  centennial 
years,  except  those  which  are   multiples  of  400.    The  act  of  Parlia- 


*  February  has  29  days  in  leap  years. 


ADVANCED   ARITHMETIC.  163 

men  t  which  changed  from  the  Julian  Calendar  (old  style)  to  the  Gre- 
gorian Calendar  (new  style)  dropped  11  days  out  of  the  month  of  Sep- 
tember, 1752,  making  September  3d,  old  style,  September  14th,  new  style. 
The  civilized  nations,  except  Russia,  have  adopted  the  Gregorian  Cal- 
endar. 

Remember  that,  when  the  number  of  a  year  is  divisible  by  4,  it  is  a 
leap  year ;  except  a  centennial  year,  whose  number  must  be  divisible  by 
400  in  order  to  be  a  leap  year. 

There  are  52  weeks  1  day  in  a  common  year,  and  52  weeks  2 
days  in  a  leap  year.  A  week  has  7  days,  named  Sunday,  Mon- 
day, Tuesday,  Wednesday,  Thursday,  Friday,  Saturday. 


EXERCISE  LXIV, 

1.  Reduce  865  da.  5  hr.  48  min.  47.8  sec.  to  seconds. 

2.  Reduce  31556986  sec.  to  higher  units. 

3.  Name  the  months  in  order  and  give  the  number  of  days  in 
each. 

Jf..  Learn  the  number  of  each  month,  as  January,  1st ;    Sep- 
tember, 9th;  June,  6th;  etc. 

5.  Find  which  of  the  following  are  leap  years :  1756,  1885, 
1800,  1892,  1876,  1900,  1904,  1906,  1952,  2000. 

6.  Find  the  number  of  days  from  June  10,  1895,  to  March 
17,  1896. 

7.  Find  the  exact  time  from  10  o'clock  a.m.,  Wednesday, 
June  12,  1901,  to  noon  the  following  Monday. 

8.  Find  the  exact  time  9170  minutes  after  8  p.  m.,  Saturday, 
June  22,  1901. 

9.  When  was  a  note  due  if  it  was  dated  June  12,  1901,  and 
due  in  98  days  ? 

10.  When  was  a  note  due  if  it  was  dated  December  10,  1879, 
and  due  in  125  days  ? 


164  ADVANCED    ARITHMETIC. 

89.  Measures  of  Value. —  The  money  of  a  country  is 
its  circulating  medium  and  adopted  standard  of  value.  The 
following  is  the  table  of  English  units  of  value,  or  denomina- 
tions of  money. 

TABIiE. 

1  pound  (£)=20  sliillings  (s.). 

1  s.  =  12  pence*  (d.). 

1  d.=4  farthings  (far.). 

Note. — The  value  of  the  English  pound  (or  sovereign)  in  United  States 
money  is  $4.8665. 

EXERCISE  XXV. 

1.  Reduce  £5  to  pence. 

2.  Reduce  7548  far.  to  £. 

3.  Reduce  8424  far.  to  higher  units. 

Jf..  A  bill  for  £200  12  s.  is  worth  how  many  dollars  ? 

5.  A  bill  for  $4300  is  worth  how  much  in  English  money  ? 

90.  Review  and  Rapid  Drill  Work.— -Heretofore 

you  have  been  required  to  consider  each  step  in  the  process  of 
reduction  of  denominate  numbers  as  a  problem,  and  to  solve  it 
as  such. 

Being  now  familiar  with  the  nature  of  reduction,  you  should 
hereafter  give  your  attention  to  speed  and  accuracy  in  obtain- 
ing the  result. 

UIRECTIONS. 

For  Reduction  Descending  :  Multiply  hy  the  number  of  units 
of  the  next  lower  denomination  which  makes  one  unit  of  the  denom- 
ination to  he  reduced. 

For  Reduction  Ascending  :  Divide  by  the  number  of  units  of 
the  given  denomination ,  which  makes  one  unit  of  the  next  higher  de- 
nomination. 

Note. — Use  the  corresponding  abstract  numbers  in  this  work. 


♦The  singular  ot pence  is  penny. 


ADVANCED    ARITHMETIC. 


165 


Reduce,  orally  — 

1.  5  bu.  to  pk. 

2.  6  qt.  to  pt. 
S.  8  pk.  to  pt. 


EXERCISE  LXVI. 


4..   14  pt.  to  qt. 

5.  48  pt.  to  pk. 

6.  128  qt.  to  bu. 


7,  48  in.  to  ft. 

8,  20  ch.  to  1. 

9,  1  mi.  to  rd. 

J3.   12  sq.  yd.  to  sq.  ft. 
14..   3  sq.  ft.  to  sq.  in. 
J5.   1  mi.  to  sq.  ch. 

W.   2  cu.  yd.  to  cu.  ft. 

20.  1  cu.  ft.  to  cu.  in. 

21,  1  bu.  to  cu.  in. 


10, 
11. 
12. 


800  1.  to  ch. 
198  ft.  to  ch. 
72  in.  to  yd. 


16.  20  A.  to  sq.  rd. 

17.  70  sq.  ch.  to  A. 

18.  480  sq.  rd.  to  A. 

22.  1  gal.  to  cu.  in. 

23.  462  cu.  in.  to  gal. 

24.  54  cu.  ft.  to  cu.  yd. 


25. 

1  day  to  min. 

26. 

1  hr.  to  sec. 

27 , 

1  yr.  to  mo. 

31. 

3  lb.  av.  to  oz. 

32. 

3  lb.  troy  to  oz. 

33. 

3  S  to  9. 

34. 

4  pwt.  to  gr. 

35. 

5  5  to  gr. 

28.  50  min.  to  hr. 

29.  6  hr.  to  da. 
80.  3  mo.  to  yr. 

36.  128  oz.  av.  to  lb. 

37.  120  oz.  troy  to  lb. 

38.  96  S  to  lb. 

39.  249  to  S. 
4.0,  120  m.  to  f  5. 


EXAMPIjSSS. 

1.  Reduce  5  bu.  to  pints. 

Process:  5x4x8x2  =  320. 
.'.  5bu.  =  320pt. 

Explanation  :    (1)  Multiplying  by  4  reduces  to  pecks ;  by 
8,  to  quarts;  by  2,  to  pints.     Therefore  the  required  number 


166  ADVANCED    ARITHMETIC, 

is  the  continued  product  of  5,  4,  8,  and  2,  or  320.  This  mul- 
tiplying should  all  be  done  mentally,  and  no  more  figures 
should  be  used  than  are  given  above. 

2.  Reduce  7  cwt.  5  lb.  4  oz.  to  ounces. 

Process :  A.  B. 

(1)  100x7  +  5  =  705.  ^  7' 

(2)  16x705+4  =  11284,        ^^^'  100 

705 
16 

4230 

705 

4 

.'o  7  cwt.  5  lb.  4  oz.  =  11284  oz.  11284 

Explanation  :  If  the  pupil  can  multiply  by  16  (and  he  ought 
to),  use  form  A  ;  if  not,  use  form  B.  In  By  the  5  is  added  to 
the  product  of  7  and  100  before  writing  the  result. 

3.  Reduce  1512  pt.  to  hogsheads. 

Process : 

A.  B. 
2)1512  3 

4)756  j^^ 

189  '  jm2^i_^. 

2  '^xix^^x^    ""• 


63)378(6  +  2  =  3 

378 


.'.  1512pt.  =  3hhd. 


Explanation  :  Form  ^  is  a  much-used  form.  Multiplying 
by  2  and  dividing  by  63  is  the  same  as  dividing  by  31|.  The 
author  has  some  preference  foi  form  B,  when  all  or  nearly 
all  of  the  divisors  are  small. 

4..  Reduce  87560  hr.  to  years. 
Process :  313 

Wm       ^313^ 
24x30x;^     72       '" 
3 
.-.  37560  hr.  =  411  yr. 


ADVANCED    ARITHMETIC.  167 


5.  Reduce  435  pt.  (dry)  to  higher  units. 

Process :     2)435 

8)217+1 

.4)27  +  1 
'6  +  3 
.-.  435  pt.  =  6  bu.  3  pk.  1  qt.  1  pt. 

6.  Express  in  integral  units  ^-f^  hhd. 

Process:     (1)    7   x^_  _ 


m 

60 

-  6U- 

21 

(2)   7  x^? 

^0x2 
20 

=W=sih 

(3)  27  x^ 

0 
10 

=  ?S=2,V 

(4)  7  y^ 

;0 

5 

=  1|. 

.-.  lb  hhd 

..=3  gal.  2qt.  1|  pt. 

7.  Reduce  .175  bu.  to  lower  integral  units 

Process  : 

.175 
4 

.7PP 
8 

5.6 
2 

1.2 

.-.  .175  bu. 

=  5qt.  1.2  pt. 

8.  Reduce  8.75  lb.  troy  to  lb.  av. 

Process : 

8.75x5760 
7000 

=  7^. 

.-.  8.75  lb. 

troy  =  7Hb.  av. 

Plan.— (1)  Multiplying  by  5760  reduces  to  gr.,  and  (2)  dividing  by  7000 
reduces  to  lb.  av. 


168                                            ADVANCED  ARITHMETIC. 

EXERCISE  LXVII. 

Reduce  : 

1.  2  mi.  20  ch.  to  inches.  7.  5  T.  4  cwt.  to  oz. 

2.  8  sq.  mi.  5  A.  to  sq.  rd.  8.   1  T.  6  oz.  to  oz. 

3.  7  cu.  yd.  5  cu.  ft.  to  cu.  in.  9.   8J  lb.  troy  to  gr. 

4.  7  bu.  5  qt.  to  pt.  10.   2  S  4  5  2  9  to  gr. 

5.  5^  gal.  1  pt.  to  gi.  11.   5  yr.  4  mo.  17  da.  to  da 

6.  5  O.  1  f  S  to  m.  12.   £7  11  d.  to  far. 

Reduce  to  higher  units : 

13.   143759  in.  17.   840  pt.  (liquid). 

IJ^.   73498  cu.  in.  18.   842  pt.  (dry). 

15.  4375  gr.  troy.  19.   546304  min. 

16.  8470  A.  20.   4444  oz.  av. 

Reduce  to  lowest  denomination  of  the  table  : 

21.  yV  yd.  25.  I  bu.  f  pk.  8  qt. 

22.  .05  yr.  26.  .5  sq.  rd.    . 

23.  -iV  rd.  27.  i  cu.  yd. +  .7  cu.  ft. 
2^.  .34  hr.  28.   .6  lb.+|  pwt. 

Reduce  to  highest  denomination  of  the  table  : 

29.  5tV  gr.  troy.  33.   15  2  9- 

30.  5pt.  (dry).  3If..  120  da. 

31.  lOOSsq.  ch.  35.  .75  qt.  (liquid). 

32.  649.  cu.  in.  36.  7  pt.  +  .8  qt. 

Reduce  : 

37.  5  lb.  av.  to  lb.  troy.  J^O.  456  gal.  to  cu.  in. 

38.  7  lb.  apoth.  to  lb.  av.  J^l.  840  bu.  to  gal. 

39.  60  bu.  to  cu.  in.  J^.  576  gal.  to  bu. 


ADVANCED    ARITHMETIC.  169 

2.     FRENCH   SYSTEM. 

91.  Linear  Measures. 

TABIiE. 

1  Myriameter  (Mm.)  =  10  Kilometers  (Km.). 

1  Km.  =  10  Hectomieters  (Ilm.). 

1  Hm.  =  10  Deliameters  (Dm.). 

1  Dm.  =  10  meters  (m.). 

1  m.  =  10  decimeters  (dm.). 

1  dm.  =  10  centimeters  (cm.). 

1  cm.  =  10  millimeters  (mm.). 

Note. — 1  meter- 39.31  inches.  By  means  of  this  equivalent,  any  num- 
ber expressed  in  units  ^of  this  table  may  be  expressed  in  units  of  the 
English  linear  table,  and  vice  versa. 

Reduction  descending  and  ascending  may  be  performed  here 
just  as  in  the  English  System ;  but,  owing  to  the  fact  that  we 
are  now  dealing  with  a  decimal  system — a  system  in  which  10 
units  of  one  denomination  make  one  of  the  next  higher — the 
process  of  reduction  may  be  much  shortened.  The  shorter 
processes  are  developed  in  the  following  examples : 

EXAMPLES. 

1.  Reduce  5  Hm.  4  Dm.  2  m.  to  m. 

Process  :  5  Hm.  4  Dm.  2m.  =542  m.,  result. 

Note. — When  there  are  no  denominations  omitted  between  the  high- 
est denomination  given  and  the  denomination  required,  the  figures 
written  in  order  side  by  side  will  be  the  abstract  number  correspond- 
ing to  the  required  result. 

2.  Reduce  7  Mm.  9  Km.  4  Dm.  to  Dm. 

Process  :  7  Mm,  9  Km.  4  Dm.  =  7904  Dm.,  result. 

Note. — Put  a  0  in  place  of  the  denomination  omitted. 

3.  Reduce  5  Km.  to  dm. 

Process  :  5  Km.  =  50000  dm.,  result. 

Note. — Put  a  0  in  place  of  each  denomination  omitted,  from  Km.  to 
dm.  inclusive. 


170  ADVANCED    ARITHMETIC. 

If.  Reduce  84579  dm.  to  higher  units. 

Process  :  34579  dm.  =  3  Km.  4  Hm.  5  Dm.  7  m.  9  dm.,  result. 

Note. — This  is  just  the  reverse  of  the  process  used  in  reduction  de- 
scending. It  is  plain,  that  by  dividing  successively  by  lO's,  the  succes- 
sive remainders  would  be  9,  7,  5,  etc.,  and  that  the  result  would  be  just 
as  given  above. 

5.  Reduce  1605  mm.  to  higher  units. 

Process  :  1605  mm.  =  1  m.  6  dm.  5  mm.,  result. 
Note.— The  0  cm.  is  omitted  in  the  result. 

6.  Reduce  5.046  Dm.  to  cm. 

Process  :  5.046  Dm.  =5046  cm.,  result. 

Note. — To  reduce  from  Dm.  to  cm. ,  we  must  multiply  by  10  three  times. 
This  is  done  by  moving  the  decimal  point  three  places  to  the  right. 

7.  Reduce  54603.25  m.  to  Hm. 

Process  :  54603.25  m.  =546.0325  Hm. 

Note. — To  divide  by  10  twice,  move  the  decimal  point  two  places  to 
the  left. 

8.  Reduce  3  Hm.  6  Dm.  to  yards. 

Process  :     (!)  3  Hm.  6  Dm.  =  360  m. 

(2)  ?80^MZ  =  393.7. 

.'.3  Hm.  6  Dm.  =393.7  yd. 
Note. — Always  reduce  to  meters^  and  then  to  inches. 

9.  Reduce  218  yd.  2  ft.  2  in.  to  Dm. 

Process  :     (1)  218  yd.  2  ft.  2  in.  =  7874  in. 

(2)        '^'^      =20. 
39.37x10 

.-.  218  yd.  2  ft.  2  in.  =  20  Dm. 

Note. — Always  reduce  to  inches^  and  then  to  meters. 

EXERCISE  LXVIII. 

Reduce  : 

1.  5  Mm.  to  m.  6.  43675  dm.  to  Km. 

2.  9  Km.  8  Dm.  to  dm.  7.  4867.3  m.  to  Hm. 

S.  7  Hm.  to  mm.  8.  4785  cm.  to  higher  units. 

4.  5  Mm.  7  Hm.  to  cm.  9.  54843  m.  to  higher  units. 

5.  8  m.  5  dm.  to  mm.  10.  4.003  Dm.  to  mm. 


ADVANCED    ARITHMETIC.  171 


11,  7  Mm.  8  Km.  6  Hm.  5  Dm.  4  m.  to  mm. 

12.  3  Mm.  8  Hm.  7  Dm.  5  dm.  4  cm.  to  mm, 
IS.  86  Mm.  to  miles. 

U.  286220  ft.  to  Dm. 

15,  640.0632  Dm.'  to. inches. 

92.  Surface  Measures. 


TABIiE. 

1  sq.  Min.  =  100  sq.  Km, 
1  sq.  Km.  =  100  sq.  Hm. 
1  sq.  HmL.  =  100  sq.  Dm. 
1  sq.  Dm.  =  100  sq.  m. 
1  sq.  m.  =  100  sq.  dm. 
1  sq.  dm.  =  100  sq.  cm. 
1  sq.  cm.  =  100  sq.  mm. 


Note. — A  square  Dekameter  is  also  called  an  are.  The  are  is  the  unit 
of  land  measure.  A  square  meter  is  equal  to  39.37x39.37  sq.  in.,  or 
about  1550  sq.  in. 

EXERCISE  LXIX. 

Reduce  : 

1.  15  sq.  Mm.  to  sq.  m.     • 

2.  75  sq.  Km.  to  sq.  dm. 

3.  4  sq.  Dm.  to  sq.  mm. 

4-.   540000  sq.  cm.  to  sq.  m. 

5.  89506000  sq.  dm.  to  ares. 

6.  48675056  sq.  Dm.  to  higher  units. 

7.  520  sq.  m.  to  sq.  in. 

8.  25  sq.  Dm.  12  sq.  m.  to  sq.  in. 

9.  17  sq.  Km.  to  sq.  yd. 

10.  520  sq.  rd.  to  sq.  m. 

11.  740  A.  to  ares. 

12.  1  sq.  mi.  to  sq.  Hm. 


172  ADVANCED    ARITHMETIC. 

93.  Solid  Measures. 

TABIiE. 

1  cu.  Mm.  =  1000  cu.  Km. 
1  cu.  Km.  =  1000  cu.  Hm. 
1  cu.  Hm.  =  1000  cu.  Dm. 
1  cu.  Dm.  =  1000  cu.  m. 
1  cu.  m.  =  1000  cu.  dm. 
1  cu.  dm.  =  1000  cu.  cm. 
1  cu.  cm.  =  1000  cu.  mm. 

Note.— A  cubic  meter  is  also  called  a  stere.  The  stere  is  used  in  meas- 
uring wood.  A  cubic  meter  is  equal  to  39.37x39.37x39.37  cu.  in.,  or 
about  60923.178  cu.  in. 

EXJ]RCISE  LXX. 

1.  75  cu.  Km.  to  cu.  m. 

2,  748  cu.  Dm.  to  cu.  cm. 

5.  10  cu.  m.  to  cu.  ft. 

4.  4856000000  cu.  dm.  to  cu.  Dm. 

6.  8000000000  cu.  m.  to  cu.  Km. 

6.  182769534  cu.  ft.  to  cu.  Dm. 

7.  How  many  cords  of  wood  in  a  pile  containing  500  cu.  m.? 

8.  How  many  steres  of  wood  in  a  pile  containing  500  cu.  yd.? 

94.  Measures  of  Capacity. 

TABIiE. 

1  Myrialiter  (Ml.)  =  10  Kiloliters  (Kl.). 

1  Kl.  =  10  Hectoliters  (HI.). 

1  HI.  =  10  Deltaliters  (Dl.). 

1  Dl.  =  10  liters  (1.). 

1  1=10  deciliters  (dl.). 

1  dl.  =  10  centiliters  (cl.). 

1  cl.  =  10  milliliters  (ml.). 

Note, — The  liter  is  equal  in  volume  to  a  cubic  centimeter  (about  1.05 
liquid  quarts,  or  .9  dry  quart). 


ADVANCED   ARITHMETIC.  173 

EXERCISE  LXXI. 

Reduce  : 

U  5  Kl,  to  1.  7.  40000  cl.  to  HI. 

2.  9  Ml.  to  DL  '..  8.  45675  dl.  to  higher  units. 

3.  7  HI.  5  1.  to  dl. '  9.  7043.08  cl.  to  Dl. 
^.  4  Kl.  4  HI.  to  cl.  10,  48  1.  to  qt. 

5.  5  Dl.  8  1.  to  qt.  11.  25  gal.  to  1. 

6.  7  HI.  to  gal.  12.  75  bu.  to  Dl. 

95 e  Measures  of  Mass. 

TABLE. 

1  Tonneau  (T.)^IO  quintals  (Q.). 

1  Q.  =  10  Myriagrams  (Mg.)» 

1  Mg.  =  10  Kilograms  (Kg.)- 

1  Kg.  =  10  Ilektograms  (Hg.). 

1  Hg.  =  10  Dekagrams  (Dg.). 

1  Dg.  =  10  grams  (g.). 

1  g.  =  10  decigrams  (dg.). 

1  dg.  =  10  centigrams  (eg.). 

1  eg.  =  10  milligrams  (mg.). 

Note. — The  gram  is  the  weight  of  a  cubic  centimeter  of  distilled  water 
(about  15.42  grains  troy). 

EXERCISE  LXXII. 

Reduce  : 

1.  6  T.  to  g.  7.   240  dg.  to  g. 

2.  9  Q.  to  Dg.  8.   4380  eg.  to  Dg. 

8.  7  Mg.  to  dg.  9.   5863  eg.  to  higher  units. 

4.  5  Dg.  to  mg.  10.   840  dg.  to  gr. 

5.  50  Mg.  to  lb.  troy.  11.   123.36  lb.  troy  to  Mg. 

6.  50  Mg.  to  lb.  av,  12.   128.36  lb.  av.  to  Mg. 


174  ADVANCED   ARITHMETIC. 

96.  Measures  of  Value. 

TABLE. 

1  franc  (fr.)  =  10  declines  (d.). 
1  d.  =  10  centimes  (c.)» 
1  c.  =  10  Hilllimes  (m.)- 

Note. — The  franc  is  equal  in  value  to  19.3i*. 

EXERCISE  LXXIII. 
Reduce : 

1.  5  fr.  to  m.  6.  $5.79  to  fr. 

2.  15  fr.  7d.  to  m.  7.  45  fr.  to  $'s. 

3.  50  fr.  to  d.  8,  8800  c.  to  /. 

4.  240  c.  to  fr.  9.  5790  m.  to  mills. 

5.  4834  m.  to  higher  units.  10,  3860  dimes  to  d.     ' 

3.     COMPOUND  DENOMINATE  NUMBERS. 

97.  Addition. 

EXAMPLES. 

1.  Add  1  bu.  2  pk.  4  qt.  1  pt. ;  5  bu.  1  pk.  2  qt. 

Process  :  Explanation  :  (1)  Write  the  names 

bu.  pk.  qt.  pt.  of  the  successive  units  of  the  table 

12      4      1  used,  putting  them  in  horizontal  or- 

der,  with  the  largest  to  the  left. 

6      3      6      1,  result.         (2)  Write    the    addends    below- 
each  part  in  its  place,  as  indicated  by  the  table  above. 

(3)  Add  as  we  have  done  heretofore,  placing  the  sum  ob- 
tained by  adding  each  column  below.  Result,  6  bu.  3  pk.  6  qt. 
Ipt. 


8gi. 


Add  2  gal.  3  qt.  1  pt.  8  gi. ;  7  gal.  1  pt.  8  gi. ;  8  gal.  2  qt. 


Explanation  :  (1)  3  gi.  +  3  gi.+3 
gi.=9  gi.  But  9  gi.=2  pt.  1  gi. 
Write  the  1  below,  and  carry  the 
2pt. 

1Q      o      n     1  1^  (2)  2pt.  (carried)  +  l  pt.  +  l  pt. 

18      3      0     l,result.  ^^^^     But4pt.=2qt.    Write  0 


Process: 

gal 

.  qt. 

pt. 

gi. 

2 

3 

1 

3 

7 

0 

1 

3 

8 

2 

0 

3 

ADVANCED    ARITHMETIC.  175 

below,  and  carry  the  2  qt.     We  use  the  0  to  show  that  there 
are  no  pints  left. 

(3)  2  qt.  (carried )  +  2  qt.  +  3  qt.  =  7  qt.  =  1  gal.  3  qt.     Write 
the  3  qt.  below,  and  carry  the  1  gal. 

(4)  1  gal.  (carried)  +  8  gal.  +  7  gal. +2  gal.  =  18  gal.     Write 
18  below.    Result,  18  gal.  3  qt.  1  gi. 

Note. — Teacher  should  point  out  the  similarity  between  this  and  Ad- 
dition of  Simple  Numbers. 

3.  Add  5  yd.  2  ft.  8  in. ;  7  yd.  2  ft.  9  Id.  ;  17  yd.  2  ft.  11  in. 

Explanation  :    (1)  Sum  28  in.  =2  ft.  4 
in.    Write  the  4  below,  and  carry  the  2. 

(2)  Sum  8  ft.  =  2  yd.  2  ft.    Write  the 
^  (ft.)  below,  and  carry  the  2  (yd.). 

(3)  Sum  31  yd.    Write  the  31  below. 
31      2      4,  result.       Result,  31  yd.  2  ft.  4  in. 


EXERCISE  LXXIV. 


Process : 

yd. 

ft. 

in. 

5 

2 

8 

7 

2 

9 

17 

2 

11 

Add: 

1.   hr.  min. 

sec. 

2.   yr.  mo.  da. 

3. 

yd. 

ft.  in. 

7  18 

86 

50  8  12 

9 

1  11 

9  85 

40 

46  5  25 

8 

0  8 

11  48 

84 

2  9  14 

11 

2  7 

J/.,   cu.  yd.  cu.  ft. 

cu.  in. 

5.   sq.  yd.  sq.  ft. 

sq. in. 

6. 

T.  cwt. 

lb. 

120   17 

540 

12     0 

78 

7  5 

40 

242    9 

1048 

46    8 

96 

18  12 

72 

488   21 

1486 

52    5 

100 

50  15 

00 

868    0 

421 

87    7 

77 

18 

84 

7.   lb.  5.  5.   9.  gr.  8.   hhd.  bbl.  gal.  qt.  pt. 

5  7  7   2  14                               2       1  25    8  1 

11  11  5    1  18                                4  0  80     2  1 

7  9  0  2  15                               5       1  11     1  0 


Process : 

T. 

15 

6 

cwt.  lb. 

6   12 

10   8 

oz. 

5 

14 

176  ADVANCED    ARITHMETIC. 

98.   Subtraction. 

EXAMPLES. 

i.  From  15  T.  6  cwt.  12  lb.  5  oz.  take  6  T.  10  cwt.  8  lb.  14  oz. 

Explanation  :  (1)  5  is  smaller 
than  14.  "We  must  take  1  lb.  from 
the  12  lb.  and  add  it  to  the  5  oz. 
1  lb. +5  oz.  =  21  oz.      21-14  =  7. 

8  16        3        7,  result.    Write  the  7  below. 

(2)  There  are  only  11  lb.  left  in 
the  minuend.    (Why?)    11-8  =  3.    Write  the  ^  below. 

(3)  10  is  larger  than  6.    (What  shall  we  do  ?)    1  T.  +6  cwt. 
=  26  cwt.    26-10=16.    Write  the  16  below. 

(4)  There  are  14  T.  left  in  the  minuend.    (Why?)    14-6 
=  8.     Write  the  8  below.    Result,  8  T.  16  civt.  3  lb.  7  oz. 

Note. — Teacher  should  point  out  the  similarity  between  this  and  Sub- 
traction of  Simple  Numbers. 

2.  A  note  dated  July  10,  1897,  was  paid  September  23,  1899. 
How  long  was  the  time  between  these  dates  ? 

Explanation  :  Write  the  latest 
date  above.    Write  the  number 
of  the    month:    September,  9th 
month;    July,   7th  month.      The 
^      2      13^  result  pupil  should  learn  the  number  of 

each  month  in  the  year. 

S.  When  I  looked  at  my  watch  last  it  was  18  min.  after  9 
o'clock,  but  it  is  now  12  min.  after  11  o'clock.     How  long  has 
it  been  since  I  looked  at  my  watch  ? 
Form  : 

j^-j*        ^2  '  Note. — 18  minutes  after  9  o'clock  is  9  hr.  18 

9  ig  min.  (from  noon  or  midnight). 

1        54,  result. 

EXEKCISE  LXXV. 

1.  From  13  bu.  1  pk.  take  8  bu.  2  pk. 

2.  From  2  cwt.  38  lb.  take  1  cwt.  49  lb. 

3.  From  2  T.  5  cwt.  41  lb.  take  1  T.  10  cwt.  74  lb. 


Process 

; 

1899 
1897 

mo. 
9 

7 

da. 
23 

10 

ADVANCED    ARITHMETIC.  177 

^.  From  5  mi.  65  ch.  1  rd.  take  3  mi.  78  eh.  3  rd. 

5.  From  50  yr.  3  mo.  12  hr.  50  min.  10  sec.  take  24  yr.  6  da. 
5  min. 

6.  From  £12  3  &.  9  d.  2  far.  take  £6  17  s.  11  d.  3  far. 

7.  From  25  cu.  yd.  16  cu.  ft.  127  cu.  in.  take  7  cu.  yd.  26  cu. 
ft.  240  cu.  in. 

8.  Find  the  time  from  Jan.  11,  1895,  to  June  1,  1901. 

99.  Multiplication. 

EXAMPIiES. 

1,  Multiply  3  bu.  1  pk.  5  qt.  1  pt.  by  7. 

Process  :  Explanation  :  (1)  7x1  pt.  =  7  pt.  = 

bu.  pk.  qt.  pt.  3  qt.  1  pt.    Write  the  1  below,  and 

3      1-^1  carry  the  3. 

(2)  7x5  qt.  =  35  qt.     35  qt.+3  qt. 

23      3      6      1,  result.     (carried)  =  38  qt.  =  4  pk.  6  qt.    Write 
the  6  below,  and  carry  the  4. 

(3)  7xlpk  =  7pk.    7pk.+4pk.  =  llpk.  =  2bu.3pk.   Write 
the  S  below,  and  carry  the  ^. 

(4)  7x3  bu.  =21  bu.    21  bu.  +  2  bu.  =  23bu.    Write  the  23 
below.     Result  J  23  bu.  3  pk.  6  qt.  1  pt. 

Note.— The  teacher  should  point  out  the  similarity  between  this  and 
Multiplication  of  Simple  Numbers. 

2.  Multiply  27  yd.  2  ft.  9  in.  by  9. 
Process:  Explanation:  (1)9x9  =  81.    81  in.  =  6 

ft.  9  in.    Write  the  9  below,  and  carry 
the  6. 

(2)  9x2=18.     18  ft.  +  6  ft.=24  ft.  =  8 
yd.    Write  0  below,  and  carry  the  8. 

(3)  9x27  =  243.     243  yd. +8  yd.  =251 
yd.    Write  the  251  below.    Result,  251  yd.  9  in. 

Multiply  5  T.  8  cwt.  54  lb.  6  oz.  by  8. 

Process : 

T.    cwt.   lb.      oz. 

5        8      54        6  Note.— Have  the  pupil  explain 

8  this  example. 

43        8      35        0,  result. 


yd. 

27 

ft.    in. 

2      9 

9 

251 

0      9,  result. 

178  ADVANCED    ARITHMETIC. 

EXERCISE  LXXVI. 
Multiply  : 

1.  12  yd.  1  ft.  10  in.  by  8. 

2.  5  mi.  43  ch.  51  ft.  9  in.  by  11. 

3.  17  sq.  yd.  7  sq.  ft.  7  sq.  in.  by  7. 

^.  8  Tp.^21  Sec.  25  A.  9  sq.  ch.  by  25. 

5.  11  cu.  yd.  16  cu.  ft.  128  cu.  in.  by  20. 

6.  24  bu.  3  qt.  by  24. 

7.  4  gal.  8  qt.  1  pt.  2  gi.  by  8. 

8.  6  Cong.  5  0.  7  f  S  B  f  5  35  m.  by  12. 

9.  7  T.  13  cwt.  46  lb.  10  oz.  by  120. 

10.  5  lb.  4  oz.  18  pwt.  22  gr.  by  9. 

11.  8  1b.  9S  7  5  29  5gr.  by  24. 

12.  5  yr.  7  mo.  3  da.  by  40. 

100.  Division. 

EXAMPIiES. 

1.  Divide  27  bu.  3  pk.  7  qt.  by  6. 

Process:  Explanation:    (1)  27  +  6  =  4,  re- 

bu.    pk.  qt.  pt.  mainder  3.    Write  the  4  below, 

6)27      3      3      0  and  reduce  the  3  bu.  to  pecks. 

4      2      4      1,  result.  (2)  3  bu.  +  3  pk.  =  15  pk.     15  +  6 

=  2,  remainder  3.    Write  the  ^ 

below,  and  reduce  the  3  pk.  to  quarts. 

(3)  3  pk.  +  3  qt.  =  27  qt.  27  +  6  =  4,  remainder  3.  Write  the 
4  below,  and  reduce  the  3  qt.  to  pints.  3  qt.  =  6  pt.  6-r-6  =  1. 
Write  the  1  below.    Result j  4  bu.  2  pk.  4  qt.  1  pt. 

Note. — Teacher  should  point  out  the  similarity  between  this  and  Divi- 
sion of  Simple  Numbers. 

2.  Divide  112  T.  16  cwt.  59  lb.  by  7. 

Process  : 

T.     cwt.    lb.  Note.— The  pupil  should  explain 

7)112      16      59  this  example. 
16       2      37,  result. 


ADVANCED    ARITHMETIC.  179 

EXERCISE  LXVII. 
Divide  : 

1.  490  bu.  2  pk.-  4  qt.  by  100. 

2.  161  yd.  1  ft.  6  in.  by  17. 

3.  3  yr.  11  mo.  4  da.  by  14. 

4.  65  gal.  2  qt.  1  pt.  by  7. 

5.  8  lb.  apoth.  by  120. 

6.  1  Tp.  5  sq.  mi.  430  A.  by  21. 

B.     INVOLUTION. 

101.  Definitions. — A  composite  number  that  can  be 
formed  by  using  the  same  number  as  a  factor  a  number  of  times 
is  called  a  PoTver  of  the  number  used ;  and  the  number  used 
is  called  the  Root  of  the  power.  The  degree  of  a  power  is  the 
number  of  times  the  root  is  used  in  producing  the  power.     Thus, 

8x3=9. 
9  is  the  second  power  of  3. 

A  small  figure  placed  just  above  and  to  the  right  of  a  num- 
ber indicates  the  degree  of  the  power  to  which  the  number  is  to 
be  raised,  and  is  called  an  Exponent.  The  number  itself  .is  the 
first  power ;  the  number  used  twice  as  a  factor  gives  the  second 
power  or  square ;  the  number  used  three  times  as  a  factor  gives 
the  third  power  or  cube  ;  used  four  times  gives  the  fourth  power ; 
five  times,  the  fifth  power,  and  so  on.     Thus, 


8^=3  (First  power.) 

8^  =  3  X  3  =  9  ( Second  power  or  square.) 
88=8X8X8=27  (Third  power  or  cube.) 
8*=8X3X3X3  =  81  (Fourth  power.) 
85=3X3X3X3X3  =  243  (Fifth  power.) 
36=3X8X8X8X8X8  =  729  (Sixth  power.) 


180  ADVANCED    ARITHMETIC. 

Involution  is  the  process  of  finding  a  required  power  of 
a  given  root. 

102.    Process.     Since   by   the   definition   a   power   is   a 
product,  tlie  process  of  finding  a  power  is  multiplication. 

EXAMPIiES. 


1. 

5*=(  )? 

Process  : 

5'*  =  5x5x5x5  = 

=  625, 

result 

2. 

m4^^{  )? 

Process  : 

534 
534 

2136 
1602 
2670 

285156, 

result. 

3, 

(A)'=(  )• 

? 

Process :  {^^f  =  i^s  X  A  x  i^?  =  j\%\j  result. 

Note. — To  raise  a  common  fraction  to  a  given  power,  both  the  nu- 
merator and  denominator  must  be  raised  to  that  power. 

4.  .048^=  (   )? 

Process  :  .0  4  8 

.0  4  3 

129 
172 


.001849 
.0  4  3 

5547 
7396 


.0  0007950  7,  result. 
Explanation :  Proceed  as  in  multiplication  of  decimals. 

EXERCISE  LXXVIII. 

Commit  to  memory  the  powers : 

i.  12;  22;  32;  4^;  5^;  6^;  7^;   8^;  9^;   10^. 

2.  P;  2^;  3^;  4^;  5^;  6^  7^  8^  9^   W. 


ADVANCED    ARITHMETIC.  181 

Raise  the  following  numbers  to  the  powers  indicated  : 

1.  171  5.   (|)«.  9,  472. 

2.  ^ 
3. 

103.  Anotlier  Metliod.  The  multiplication  may  be 
performed  by  separating  the  root  into  two  numbers,  and  mul- 
tiplying each  part  of  the  multiplicand  by  each  part  of  the 
multiplier. 

BXAMFLSS. 


1272. 

K      6,  iuV' 

10,  mv\ 

87^. 

■>  7.  (mr- 

11,  5.062. 

13^ 

8.   (24&i)^ 

1^>  (.141) 

1. 

52z 

=  (8+2)'^ 

=  (    )? 

Process  : 

3+2 
3+2 

6+4 
9+  6 

9+12+4  =  25, 

2. 

T-- 

={6+2y 

=  (    )? 

Process : 

5+2 

5  +  2 

10  +  22 
52  +  10 

52+2x10+22  =  25+20+4  =  49,  result. 
These  examples  are  illustrative  of  the  following  principle : 

Principle  :  The  square  of  the  sum  of  two  numbers  is  equal  to 
the  square  of  the  first  plus  twice  the  product  of  the  first  times  the 
second  plus  the  square  of  the  second. 

By  referring  to  the  two  numbers  as  1st  and  2d,  this  principle 
may  be  stated  in  an  equation  as  follows : 

(lst+2cl)2=lst2+2xlstX2cl  +  2d^  (Commit.) 


182 


ADVANCED    ARITHMETIC. 


62=.(    )? 

Process 


62  =  (4  +  2)2  =  42+2x4x2+22=  16  +  16+4  =  36;  q^, 
62  =  (3+3)2  =  32  +  2x3x3+32  =  9  +  18+9  =  36;  or, 
62  =  (l+5)2  =  12+2x1x5+52  =  1  +  10+25  =  36,  result. 


Note. — A  number  may  be  separated  into  any  two  of  its  addends  and 
this  principle  will  apply. 

i.    (17)^=(  )? 

Process:  17='  =  (lO+7)2  =  102+2xlOx7+72  =  10O+140+49  =  289,  result. 


S. 


5'=(    )? 

Process  : 


6.    9«=(   )? 
Process . 


2+  3 
2+  3 

6+  9 
4+  6 

4+12+  9 
2+  3 

12+36+27 

8  +  24  +  18 

8+36  +  54+27  =  125,  result. 


5+4 
5+4- 


5x4+42 
52+5x4 

52+2x5x4+42 

5  +  4 

52x4+2x5x42+4'" 
5^2x52x4+       5x42 

53+3x52x4+3x5x42+4^  =  729,  result. 

Note. — The  teacher  should  carefully  explain  each  step  in  this  multi- 
plication. 

This  example  illustrates  the  following  principle : 

Principle  :  The  cube  of  the  sum  of  tivo  numbers  is  equal  to  the 
cube  of  the  first  plus  three  times  the  square  of  the  first  times  the  sec- 
Olid  "plus  three  times  the  first  times  the  square  of  the  second  plus  the 
cube  of  the  second  ;  or. 


ADVANCED   ARITHMETIC.  183 

(lst+2cl)^=  lst^+3  X Ist^X  2(1  +  3  X  1st  X  2d2  +  2dK 

(Commit.) 

7.    233=(  )?    J; 

Process:  23'^  =  ^20+3)3  =  203+3x202x3+3x20x3«+33  = 
8000 + 3600 + 540 + 27  =  12167,  result. 

Note.— These  two  principles  have  been  here  presented,  not  so  much 
for  their  practical  value  in  involution,  as  to  prepare  for  their  use  in  evo- 
lution— extracting  square  root  and  cube  root. 

EXERCISE  LXXIX. 

1.  lP=:(7+4)2=(   )?  6.  83=(3+5)«  =  (   )? 

2.  102=(3+7)2=(  )?  7.  123=(10+2)3=(   )? 

3.  132=(5+8)2=(  )?  8.  143=(10+4)3=(  )? 
.4.  212=(20+1)2=.(   )?  9,  233=(20+3)3=(   )? 

S.   342= (30+4)2=  (  )^  2Q^   46^=  (40+6)3=:  {  )? 

0.    EVOLUTION. 

104.  Definition. — Evolution  is  the  process  of  find- 
ing a  required  root  of  a  given  power. 

The  number  itself  is  the  first  root ;  one  of  the  two  equal  fac- 
tors which  compose  a  number  is  its  second  ov  square  root ;  one  of 
the  three  equal  factors  which  compose  a  number  is  its  third  or 
cube  root;  one  of  the  four  equal  factors  which  compose  a  num- 
ber is  its  fourth  root;  and  so  on. 

The  required  root  of  a  given  number  may  be  indicated  in 
either  of  two  ways : 

( 1 )  By  the  use  of  the  radical  sign,  t/,  with  a  small  figure  or 
figures  placed  between  the  two  parts  to  indicate  the  degree  of 
the  root.     Thus, 

-^121,  the  square  root  of  121. 
-Vrm,  the  cube  root  of  1728. 
1^^256,  the  fourth  root  of  256. 


184  ADVANCED    ARITHMETIC. 

(2)  By  a  fractional  exponent.     Thus, 

121^,  the  square  root  of  121. 

1728^,  the  cube  root  of  1728. 

256^,  the  fourth  root  of  256. 

Note. — The  radical  sign  without  the  small  2  is  commonly  used  to  in- 
dicate the  square  root  of  a  number.     Thus, 

"V  64,  the  square  root  of  64. 

1.     SQXJABE  HOOT. 

105.  First  Process. —  The  square  roots  of  integral 
squares  can  usually  be  obtained  by  factoring.  When  a  num- 
ber is  separated  into  two  equal  factors,  one  of  the  factors  is 
the  square  root  of  the  number. 

EXAMPLES. 

1.  -i/225,-:(   )  ? 

Process:  225  =  5x5x3x3  =  15x15. 
•••  -v/225  =  15,  result. 


2,  -i/l44x64=(   )  ? 

Process  :  144  x  64=  12  x  12  x  8  x  8  =  96  x  96. 
•'•  -1/144x64  =  96,  result.^ 

EXERCISE  LXXX. 

1.  Commit  to  memory  the  square  root   ^f  1,  4,  9,  16,  25,  36, 
49,  64,  81,  100. 

Find  by  factoring  the  square  root  of — 

2.  900  5. '824  8.  81x25 

.       3.  625  6.  576  9.  676x100 

4.  256  7.  1296  10.  144x729 

106.    Second   Process. — When  numbers  are  large,  or 
when  they  are  not  perfect  squares,  the  process  to  be  employed 


ADVANCED    ARITHMETIC.  185 

in  extracting  the  square  root  is  developed  from  the  following 
equation,  which  was  learned  in  Involution: 

(lst-|-2d)^  =  lst'+2xlstx2d+2d^ 

Preparatory  to  the  development  of  this  process,  the  following 
principles  should  be  learned : 

Principles:  1.  When  an  integral  number  is  separated,  from 
right  to  left,  into  periods  of  two  figures  each,  the  number  of  periods 
thus  formed  will  equal  the  number  of  integral  places  in  the  square 
root  of  the  number. 

Note, — This  principle  holds,  even  though  there  may  be  but  one  figure 
in  the  left-hand  period. 

This  principle  may  be  illustrated  as  follows : 

-1/1==  1, 

T/roo==io, 

ViWob=ioo, 
Vroowoo^iooo. 

Thus,  it  appears  that  for  integers  less  than  l^'OO,  the  square  root  can- 
not have  more  than  one  integral  place ;  for  integers  less  than  l^OO'OO, 
the  square  root  cannot  have  more  than  two  integral  places;  and  so  on. 

^.  When  an  integer  is  separated  from  right  to  left  into  periods  of 
two  figures  each,  the  square  root  of  the  largest  integral  square  in  the 
left  period  will  give  the  first  or  left  figure  of  the  square  root  of  the 
number ;  the  square  root  of  the  largest  integral  square  of  the  number 
formed  by  the  first  two  periods  will  give  the  first  two  figures  of  the 
square  root  of  the  number ;  and  so  on. 

This  principle  may  be  illustrated  as  follows : 


2862=55696;  then, -i/5'56'96= 236. 
The  largest  integral  square  in  5  is  4. 
1/^4  =  2. 


186  ADVANCED    ARITHMETIC. 

Then,  2  is  the  first  figure  of  the  square  root,  236. 
The  largest  integral  square  in  556  is  529. 

-^-"529  =  23. 
Then,  3  is  the  next  figure  of  the  square  root,  236. 

'  EXAMPLE. 

Find  the  square  root  of  288869. 

Step  1. — To  find  the  first  or  left  figure  of  the  root. 
Process  :  Explanation  :    By  separating  the  number 

28^83^6915  into  periods  of  two  figures  each,  it  appears 

25  that  the  square  root  will  contain  three  fig- 

3  ures  (Prin.  1).     The  largest  integral  square  in 

28  is  25,  of  which  5  is  the  square  root.    There- 
fore, the  first  or  hundreds  figure  of  the  root  is  5  (Prin.  2). 

Step  2. — To  find  the  second  figure  of  the  root.  By  Step  1,  it 
was  found  that  the  first  figure  of  the  root  is  5 ;  then,  the 
number  represented  by  the  first  two  figures  of  the  root 
must  lie  between  50  and  60,  and  the  square  of  this  number  is 
the  largest  integral  square  in  2883,  the  first  two  periods  of 
the  number  whose  root  is  to  be  found  ( Prin.  2).  If  this  num- 
ber, which  lies  between  50  and  60,  be  represented  as  the  sum 
of  two  numbers,  then  its  square  may  be  represented  by  the 
following  equation : 

(Ist+2d)2  =  lst2+2xlstx2d+2d2;  or, 
(lst+2d)2=  lst2+2d(2  X  lst+2d) 

Now,  since  (1st + 2d)  represents  a  number  lying  between  50 

and  60,  let 

1st  =  50, 

and  proceed  to  find  2d. 

Process  :  28^83^69 1 50 + 3  =  53. 

Ist2  =  2500  (to  be  subtracted) 25  00      '' 


2x lst  =  100  (trial  divisor) 100 

2d  =  3  (found  by  trial). 3 

2  X  1st + 2d  =  103  (complete  divisor) ...     103 
2d(2 X  1st  +  2d)  =  309  (to  be  subtracted) 


383 

309 

74 


ADVANCED   ARITHMETIC. 


187 


Explanation  :  After  subtracting  the  2500  from  the  first  two 
periods,  there  is  left  383.  Now,  383  must  contain  100  x  2d+2d^. 
Since  the  2d  represents  tlie  units  of  the  second  place,  2d^  is 
small  as  compured  to  100  xM;  finding  how  many  times  383 
contains  100  will'  practically  find  the  2d.  Using  100  for  a 
trial  divisor,  proceed  just  as  in  finding  the  quotient  figure  in 
long  division.  The  2d  is  found  to  be  3.  The  complete  divisor 
is  formed  by  adding  the  3  to  the  100,  making  103.  This  com- 
plete divisor  multiplied  by  the  3  in  the  root  gives  309,  the 
number  to  be  subtracted. 

Step  3. — To  find  the  third  or  units  figure  of  the  root.  By 
Steps  1  and  2,  the  first  and  second  figures  of  the  root  are  5 
and  3  respectively.  Then,  the  number  represented  by  the 
three  figures  of  the  root  must  lie  between  530  and  540,  and 
the  square  of  this  number  must  not  be  greater  than  288369. 
As  in  Step  2,  consider  this  root  as  made  up  of  two  parts, 
and  its  square  as  represented  by  the  equation — 

(I8t+2d)2  =  l8t2+2d(2xlst+2d). 

Let  1st  =  530, 

and  proceed  to  find  the  2d  as  before. 


Process : 
lst2  =  280900. 


28^83^69 1 530 +7 
28  09  00 


537,  result. 


2x  lst  =  1060 1060 

2d  =  7 7 

2xlst  +  2d  =  1067 1067 

2d(2xlst+2d)  =  7469 


74  69 


7469 


EXERCISE  LXXXI. 


By  steps  as  above,  find  : 
1.  -i/l849 


2.  -1/19044 


S.  -1/478864 
4.  -i/950625 


107.     Second    Process    Shortened. —  The  process 
of  extracting  square  root,  as  just  given,  may  be  much  shortened 


188 


ADVANCED    ARITHMETIC. 


(1)  by  consolidating  the  several  steps  into  one  continuous  pro- 
cess. In  example,  p.  186,  25  subtracted  in  Step  1  amounts  to  the 
same  as  the  2500  subtracted  in  Step  2,  and  the  280900  sub- 
tracted in  Step  3  amounts  to  the  same  as  the  2500  and  the  809 
subtracted  in  Step  2;  (2)  by  dropping  all  O's  that  do  not  affect 
results. 


EXAMFIiSS. 


1.   -1/288369  =(   )  ? 


Process 


28^83^69  537 
25  ' 


103 


383 

309 


Explanation:  (1)  Separate  the  number 
as  before,  into  periods  of  two  figures  each. 

(2)  5x5  =  25.  Place  5  in  the  root,  sub- 
tract 25  from  28,  and  bring  down  the 
next  period. 

(3)  2x5  =  10.  Place  10  on  the  left  for  a 
trial  divisor,  and  try  10  into  38  (instead  of 
100  into  383  as  before).  38  +  10  =  3  +  .  Then, 

3  is  the  second  figure  of  the  root.  Place  the  3  on  the  right 
of  the  5  in  the  root  and  on  the  right  of  the  10  in  the  divisor. 
Multiply  103  by  3,  write  the  product  309  beneath  the  383,  sub- 
tract, and  bring  down  the  next  period. 

(4)  2x53=  106.   Place  the  106  on  the  left  for  atrial  divisor, 
and  proceed  as  before. 


1067  7469 
7469 


Note. — The  teacher  should  assist  in  comparing  this  shortened  process 
with  the  process  by  steps,  and  explain  all  differences  not  clear  to  the 
pupil. 


-1/611.5729  =  (   )  ? 


Process  : 

6^1.57^29124.73 

4 ' 

441211 
1176 

48713557 
13409 

4943 


14829 
14829 


Note. — In  decimals,  the  periods  are 
formed  from  the  decimal  point  to  the 
right.  Otherwise,  the  process  is  the 
same  as  in  integers. 

When  a  number  is  not  an  exact 
power,  O's  may  be  added  for  decimal 
places,  and  the  process  continued  at 
will. 


ADVANCED    ARITHMETIC. 


189 


3.   -i/ 5798649  =(   )  ? 

Process  : 


5^79^36^49 1 2407,  result . 


44 


179 
176 


4807 


33649 
33649 


Explanation  :  If,  after  bringing 
down  a  period,  the  number  thus 
formed  is  too  small  to  contain 
the  trial  divisor,  place  a  0  in  the 
root,  bring  down  the  next  period, 
cancel  the  old  trial  divisor,  form 
a  new  trial  divisor,  and  proceed 
as  before. 


4.  V.iWi={  )?      _ 

Process:    (1)  Vq25  =  V 25x25  =  25 
(2)  Vl024  =  -i/32x32  =  32 
.'.  ^^=11,  result. 

Explanation :  The  square  root  of  a  common  fraction  is  the  square 
root  of  its  numerator  divided  by  the  square  root  of  its  denominator. 
This  follows  as  a  reverse  of  the  process  of  squaring  a  common  fraction.  Com- 
mon fractions  whose  terms  are  not  integral  squares,  should  be  reduced 
to  decimals,  and  their  square  roots  should  be  extracted  as  in  No.  5, 
below. 


5.   t/.048=(   )?     (Result  true  to  4  decimal  places.) 

.04^30^00^00 1. 2073 +  ,  result.  Explanation:   (1)  When  the 

right-hand  period  of  a  decimal 
is  not  complete,  it  should  be 
made  so  by  annexing  O's. 

(2)  When  there  is  a  remain- 
der after  using  the  last  period 


407 
4143 


3000 
2849 


15100 
12429 


in  either  decimals  or  integers,  periods  of  O's  may  be  annexed 
and  the  process  continued  at  will. 


EXERCISE  LXXXII. 


Find  the  value  of- 


1. 

2. 
3. 

i/6889 
-i/5625 
t/8186 
t/8025 

5. 
6. 

7. 
8. 

t/ 62001 
T/98G86 
-1/4556.25 

9. 
10. 
11. 
12. 

1/^.1406844881 

4. 

-1/16.096144 

(tVA¥9)^ 

190  ADVANCED    ARITHMETIC. 

Obtain  results  true  to  three  decimal  places  : 

IS.    2^                      17.   t/8.  141592  U.  (|)^ 

U-    B^                      18.   -i/4b?i  22.  (424)^ 

15.    5^                      19.    -]/ 326.004  ^^.  (28111)^ 

i^.    11^                   20.   t/.OOOT  ^^.  (33.47^)^ 

2.     CUBE   BOOT. 

108.  First  Process. — The  cube  roots  of  integral  cubes 
can  usually  be  obtained  by  factoring.  When  a  number  is  sep- 
arated into  three  equal  factors,  one  of  the  factors  is  the  cube 
root  of  the  number. 

EXAMFIiES. 


1.    #^1728=  (   )? 

Process  :  1728  =  4  x  4  x  4  x  3  x  3  X  3  =  12  x  12  x  12. 
.-.  ^1728  =  12,  result. 


2.    1^64 X 512  =(   )? 

Process  :  64x512  =  4x4x4x4x4x4x2x2x2  =  32x32x32. 
.'.    #"64x512  =  32,  result. 

EXERCISE  LXXXIII. 

1.  Commit  to  memory  the  cubes  of  the  integers  from  1  to  10. 
By  factoring  f  find  — 

2.  1^2744. 

s.  fimi. 

4.    #^4851. 

109.  Second  Process. — When  numbers  are  large  or 
when  they  are  not  perfect  cubes,  the  process  to  be  employed  in 
extracting  the  cube  root  is  developed  from  the  following  equa- 
tion, which  was  learned  in  Involution : 

(Ist+2d)3=lst^+3xlst2x2d+3xlstx2d2+2d^ 


5.    #"3375. 

8.    #"27X729. 

6.    1^4096. 

9.    #"216X1000. 

7.    1^8x729. 

10.    #"512X343. 

ADVANCED   ARITHMETIC.  191 

Preparatory  to  the  development  of  the  process,  the  following 
principles  should  be  learned  : 

Principles  :  it^When  an  integral  number  is  separated  from 
right  to  left  into  periods  of  three  figures  each,  the  number  of  periods 
thus  formed  will  equal  the  number  of  integral  places  in  the  cube  root 
of  the  number. 

Note. — This  principle  holds  even  though  there  may  be  but  one  or  two 
figures  in  the  left-hand  period. 

To  illustrate : 

n=i. 

#'Pooo=io. 
f^rooo'ooo^ioo. 
#^rooo^ooo^ooo= 1000. 

Thus,  it  appears  that  for  integers  less  than  1^000,  the  cube  root  can- 
not have  more  than  one  integral  place ;  for  integers  less  than  1^000^000 
the  cube  root  cannot  have  more  than  two  integral  places ;  and  so  on. 

2.  When  an  integer  is  separated  from  right  to  left  into  periods 
of  three  figures  each,  the  cube  root  of  the  largest  integral  cube  in  the 
left  period  will  give  the  first  or  left  figure  of  the  cube  root  of  the  num- 
ber ;  the  cube  root  of  the  largest  integral  cube  of  the  number  formed 
by  the  first  two  periods  will  give  the  first  two  figures  of  the  cube  root 
of  the  number;  and  so  on. 

To  illustrate : 


236^=:  13^144^256;  then,  1^13^44^256=236. 
The  largest  integral  cube  in  13  is  8. 

1^8  =  2. 
Then,  2  is  the  first  figure  of  the  cube  root,  236. 
The  largest  integral  cube  in  13144  is  12167. 


f  12167  =  23. 
Then,  3  is  the  second  figure  of  the  root,  236. 


192  ADVANCED    ARITHMETIC. 

EXAMPIiE. 

Find  the  cube  root  of  154854158. 

Step  1. — To  find  the  first  or  left  figure  of  the  root. 

Process  :   154^854^53(5 
125 
29 
Explanation  :   By  separating  the  number  into  periods  of 
three  figures  each,  it  appears  that  the  cube  root  will  contain 
three  figures  (Prin.  1. )    The  largest  integral  cube  in  154  is  125, 
of  which  5  is  the  cube  root.    Therefore,  the  first  or  hun- 
dreds figure  of  the  root  is  5  (Prin.  2). 

Step  2. — To  find  the  second  figure  of  the  root.  By  step  1  it 
was  found  that  the  first  figure  of  the  root  is  5 ;  then  the  num- 
ber represented  by  the  first  two  figures  of  the  root  must  lie 
between  50  and  60,  and  the  cube  of  this  number  is  the  larg- 
est integral  cube  in  154854  (Prin.  2). 

If  this  number  which  lies  between  50  and  60  be  repre- 
sented as  the  sum  of  two  numbers,  then  its  cube  may  be  rep- 
resented by  the  following  equation : 

(Ist  +  2d)3  =  lst3+3xlst2x2d+3xlstx2d2+2d3;  or, 
(Ist+2d)3  =  lst3+2d(3xlst2+3xlstx2d+2d2). 

Now,  since  (lst+2d)  represents  a  number  lying  between 

50  and  60,  let 

1st  =  50, 

and  proceed  to  find  2d. 

Process  :                                                            154^854^53 1 50+3  =  53. 
Ist3= 125000  (to  be  subtracted) 125000         ' 

3xlst2  =  7500  (Trial  divisor) 7500~ 

3  x  1st  X  2d  =  450 450 

2d2  =  9 9 


3  X  lst2+3  X  1st  X  2d+2d2  (Complete  divisor)    7959 
2d(3  X  lst2+3  X  1st  X  2d  +  2d2) 


29854 


23877 


5977 


Explanation  :  After  subtracting  the  125,000  from  the  first 
two  periods,  there  is  left  29854.  Now,  29854  must  contain 
M{3xlst^-\-Sxlstx2d-\-2d^).  Of  this  expression,  the  2d  repre- 
sents the  figure  of  the  root  sought,  and  {3xl8t^-\-3xlstxM-{- 
M^)  represents  the  complete  divisor.    Since  the  M  represents 


ADVANCED   ARITHMETIC.  193 

the  units  of  the  second  place  in  the  root,  the  second  and 
third  terms  of  the  complete  divisor  are  small  as  compared 
with  the  first.    Therefore,  Sxlst^,  or  7500,  is  used  as  a  trial  ^ 

divisor.  Finding  the  root  figure  is  done  in  the  same  way  as 
finding  the  quotient  figure  in  long  division ;  but  since  the 
trial  divisor  is  somewhat  smaller  than  the  complete  divisor, 
allowance  must  be  made  for  this  in  selecting  the  figure  of 
the  root.  The  second  figure  of  the  root  is  found  to  be  3. 
The  complete  divisor  is  found  by  adding  to  the  7500  the  values 
of  5  X  1st  X  M  and  2d^,  or  450  and  9,  making  in  all  7959.  This 
complete  divisor,  multiplied  by  the  root  figure  3,  gives  the 
number  to  be  subtracted,  23877. 

Step  3. — To  find  the  third  figure  of  the  root.  The  first  and 
second  figures  of  the  root  have  already  been  found  to  be  5 
and  3  respectively.  Then,  the  number  represented  by  the 
three  figures  for  the  root  must  lie  between  530  and  540,  and 
the  cube  of  this  number  must  not  be  greater  than  154854153. 
As  in  Step  2,  consider  this  root  as  made  up  of  two  parts  and 
its  cube  as  represented  by  the  equation, 

(lst+2d)3=  lstH2d(3  X  lst2+3  X  1st  x2d+2d2). 

Let  1st  =  530, 

and  proceed  to  find  the  M  as  before. 

154'854a53|530+7  =  537,  result. 
Ist3  =  148877000 148877000  ' 

3  X  lst2  =  842700 842700 

3  X  1st  x2d  =  11130 11130 

2d2  =  49 49 


3  X  lst2+3  X  1st  X  2d  +2d2 853879 

2d(3  X  lst2+3x  1st  x2d  +  2d2) 


5977153 


5977153 


EXERCISE   LXXXIV. 

By  steps  as  above,  find — 


1.  f^274625.  3.    #^94818816. 

2.  ^^7189057.  4.    f2087336952. 


194 


ADVANCED    ARITHMETIC. 


110.  Second  Process  Sliortened. — The  process  of 
extracting  cube  root,  as  just  given,  may  be  much  shortened 
( 1 )  by  consolidating  the  several  steps  into  one  continuous  pro- 
cess and  (2)  by  dropping  all  O's  that  do  not  affect  the  result- 


EXAMPXiES. 


#"  154854153 

Process : 


(    ) 


154^854^53  537,  result. 
125  


7500 
450 


29854 


7959 

842700 
11130 
49 

853879 


23877 


Explanation:  (1)  Separate 
the  number  as  before  into 
periods  of  three  figures  each. 
(2)53=125.  Place  the  5  in 
the  root,  subtract  the  125 
from  154,  and  bring  down 
the  next  period. 

(3)  The  trial  divisor  is  formed 
by  annexing  a  0  to  the  root  al- 
ready found,  squaring  the  re- 
sult and  multiplying  the  square 
by  3.  502x3  =  7500.  29854  + 
7500  =  3  +  .  Place  the  3  in  the  root.  The  complete  divisor  is 
obtained  by  adding  together  the  trial  divisor,  three  times  the  last 
figure  of  the  root  times  the  remainder  of  the  root  with  a  0  an- 
nexed, and  the  square  of  the  last  figure  of  the  root.  3  x  3  x  50  = 
450;  32  =  9.  7500+450+9  =  7959,  complete  divisor.  Multiply 
the  complete  divisor  by  the  last  figure  of  the  root,  subtract 
the  product,  and  bring  down  the  next  period. 
(4)  Proceed  as  in  (3). 


5977153 


5977153 


r8489664=(  )  ? 

Process  : 

8^489^664 1 204 
8  ' — 


120000 

2400 

16 


122416 


489664 


489664 


Explanation  :  If  after  bringing 
down  a  period,  the  number  thus 
formed  is  too  small  to  contain  the 
trial  divisor,  place  a  0  in  the  root, 
bring  down  the  next  period,  can- 
cel the  old  trial  divisor,  form  a 
new  trial  divisor,  and  proceed  as 
before. 


Process  :  #"? |f=  |,  result. 


ADVANCED    ARITHMETIC. 


195 


Explanation  :  The  cube  root  of  a  common  fraction  is  equal 
to  the  cube  root  of  the  numerator  divided  by  the  cube  root 
of  the  denominator.  Common  fractions,  whose  terms  are  not 
integral  cubes,  should  be  reduced  to  decimals,  and  their 
cube  roots  should  be  extracted  as  in  No.  4  or  5  below. 


4.    r  .074088=  (   )  ? 
Process  : 


.074^088  .42,  result. 
64        ' 


4800 

240 

4 


5044 


10088 


10088 


Explanation :  Separate 
the  number  from  the  decimal 
point  to  the  right  into  periods 
of  three  figures  each  ;  then, 
proceed  as  in  integers. 


5.    ri2.8458: 

Process  , 


(   )  ?     (Result  true  to  three  decimal  places.) 


12^345'800'000  2.311  +  ,result. 


1200 

180 

9 

4345 

1389 

4167 

1587( 
6J 

X) 
)0 

1 

178800 

159391 

159391 

160083C 
69S 

K) 
10 

1 

19409000 

m)V 

)2f 

n 

16015231 

Explanation  :  (1)  When  the 
right-hand  period  of  a  decimal 
is  not  complete,  it  should  be 
made  so  by  annexing  O's.  (2) 
When  there  is  a  remainder 
after  using  the  last  period  in 
either  decimals  or  integers, 
periods  of  O's  may  be  annexed 
and  the  process  continued  at 
will. 


3393769 


Find  the  value  of  : 

1.  f^l5625 

2.  1^18824 


3.  #"1225043 


EXERCISE  LXXXV. 


^.  ^517781627 
5.  #"189119224 


6.  r 1967221277 


196  ADVANCED    ARITHMETIC, 


7.  r65804767688  U.  f^66485045871501 

8,  #^997002999  15.  ^1225.043 


9.  #^260917119  16.  #"529.475129 

10.  f^3422470843  17.  f/.013312053 

11.  f/351895816000  18.   (tj^'VV)^" 

12.  #^35778897375  19.  {tUU)^ 

13.  1^1881242513625  20.  (tWA^)^ 

Obtain  results  true  to  three  decimal  places  : 

21.  2^  23.  3.6^  25.  .006^ 

22.  4^  2^.  .244^  26.   (247|)^ 

3.     BOOTS  OF  SIGETER  DEGREES. 

111.  Process. —  If  the  index  of  a  root  has  no  other  fac- 
tors than  2's  or  S's,  that  root  may  be  extracted  by  applying 
processes  already  explained. 

Principle:  Any  root  may  he  extracted  hy  extracting  in  succes- 
sion the  roots  indicated  hy  the  factors  of  the  index  of  that  root. 

Explanation  :  The  4th  root  is  the  square  root  of  the  square 
root ;  the  6th  root  is  the  square  root  of  the  cube  root ;  the 
8th  root  is  the  square  root  of  the  square  of  the  square  root ; 
and  so  on. 

EXAMPLES. 

1.    -1^64=  (   )  ? 

Process:   (1)  ^64  =  4. 


(2)  -y  4  =  2,  result. 

Note.— The  order  is  not  material.     The  square  root  may  be  extracted 
first,  and  then  the  cube  root. 


2.    '1^531441=  (   )  ? 

Process:   (1)  #"531441  =  81. 
{2)V8i  =  d. 
(3)  Vg  =  3,  result. 


ADVANCED   ARITHMETIC.  197 

Roots  may  often  be  found  by  factoring  and  inspecting  the 
result. 

3,   -^3125=  (   )  ? 

Process:   3125  =  5x5x5x5x5. 
.'.-^^3125  =  5,  result. 


4.     yi401=(    )  ? 

Process  :   1401  =  7x7x7x7. 

.'.  -v^iiol  =  7,  result. 


EXERCISE  LXXXVI. 


1.  -i5^256  3.  -^^40353607  5.  -if^78125 


2.  -iJ^  14641  4.    1^531441  6.  "1^248832 

D.     NUMBERS  EXPRESSED  BY  WORDS  OR  LETTERS. 

112.  Explanations. —  In  writing  out  the  solutions  of 
problems,  a  great  many  ivords  are  used  to  express  numbers. 
Some  of  the  words  commonly  used  in  this  way  are  the  follow- 
ing: 

number  area 

value  volume 

price  work 

cost  force 

gain  time 

loss  interest 

weight  principal 

length  rate 

width  sum 

thickness  amount,  etc. 

NoTB.— Suppose  this  problem  is  given:  "Five  times  the  cost  of  my 
horse  is  $320;  find  its  cost."  The  relation  given  in  the  problem  may  be 
properly  expressed  as  follows : 

5  X  cost  =  $320. 


198  ADVANCED   ARITHMETIC. 

Here  the  word  "  cost  "  refers  to  a  particular  number  and  *'  5  x  cost  "  is  5 
times  that  number. 

Not  only  are  words  used  to  express  numbers,  but  letters  are 
also  thus  used. 

Note. —  Suppose  that 

Cost  of  1  apple  =  If ;  then 

Cost  of  2  apples  =  2^  ; 

Cost  of  3  apples  =  3^ ;  and  in  general, 

Cost  of  any  number  of  apples  =  that  number  of  cents. 
Now,  if  n  be  used  for  the  word  number,  the  expression  becomes — 

Cost  of  n  apples  =  n^. 

In  the  same  way  it  is  common  to  use  v  for  value,  p  for  price, 
c   for  cost,  etc. 

It  is  as  necessary  to  add,  subtract,  multiply,  and  divide  num- 
bers thus  expressed  as  it  is  to  perform  these  processes  with 
numbers  expressed  in  figures. 

113.   Addition. 

BXAMPI.E8. 


1. 

2. 

3. 

4. 

5. 

6. 

40 

4X10 

4  tens 

4Xco8t 

4Xc 

4:C 

50 

5x10 

5  tens 

5  X  cost 

5Xc 

5c 

60 

6X10 

Qtens 

QXcost 

6xc 

6c 

150,  sum.  15  X  10,  sum.  15  tenS,  sum.  15  X  cost,  sum.  15  X  C,  sum.  15  C,  sum. 

Note. — Carefully  compare  the  above  examples  (1)  as  to  form  of  ex- 
pression and  (2)  as  to  results.  Notice  that  ''4c  "  is  read  four  c,  and  means 
4  c's  or  4XC.    As  a  number,  4c  is  the  same  as  ^ x c. 

7.     Add  lOXcost,  7xcost,  bXnumber,  9Xcost,  and  QXnumber. 

Process  :   10  x  cost + 5  x  number 
7  X  cost +Qx  number 

9  x  cost 

26  X  cost  + 11 X  number  J  result. 

Note. — In  this  example  there  are  five  numbers  to  be  added,  three  of 
one  kind  and  two  of  another.    Those  of  the  same  kind  are  united.    The 


ADVANCED   ARITHMETIC.  199 

resulting  numbers  cannot  be  united  into  one  number ;  but  the  sign,  + , 
placed  between  them,  shows  that  they  are  to  be  considered  together  as 
one  sum. 


^K. 

EXERCISE  LXXXVn. 

Add: 

1.    . 

2. 

3. 

^. 

5  X  weight 

17  X  area 

3  X  time 

mxioss 

7  X  weight 

21 X  area 

12  X  time 

loss 

9  X  weight 

area 

21 X  time 

12  X  loss 

15  X  weight 

11 X  area 

time 

27  Xloss 

5. 

6. 

7. 

8 

9. 

IBXg 

75Xc 

12  Xamt. 

53  X/ 

ISXth. 

7xg 

ISXc 

96  X  ami. 

121  X/ 

72  xth. 

25Xg 

SXc 

S7Xamt. 

137  X/ 

128  xth. 

ISXg 

54Xc 

19  Xamt. 

86  X/ 

94Sxth. 

20Xg 

75Xc 

96  Xamt. 

243  X/ 

th. 

10. 

11. 

12. 

13. 

u. 

72  r 

85  Z 

18  w; 

7v 

248 

Sr 

37 « 

mw 

89 1; 

178 

148  r 

128? 

172  w 

27 1; 

12  8 

59  r 

374  i 

94  ty 

34  V 

98 

15.  12 X work,  17 X work,  9Xwork,  7Xsum,  18 X sum,  14Xwork, 
2QXsum,  QXwork,  34 X sum,  and  12 X sum. 

16.  5Xc,  7X/,  8Xc,  12Xc,  17x/,  18x/,  and9Xc. 

17.  24  w,  72  w,  120  w,  74 1,  95  t,  71  w,  and  39  t. 


200  ADVANCED    ARITHMETIC. 

114.    Subtraction. 


EXAMPIiEB. 


i. 

25  X  value 
17  X  value 
Sx  value,  dif. 


25X^7 
17X^; 
8  X 17,  dif. 


25  V 

17i? 
8'y,dif. 


4. 
25p 
17p 
8^,  dif. 


5.  From  25  c  take  17  n. 


Process 
25c 
17  n 


25c-17n,  result. 


Note. — The  minuend  and  subtrahend 
are  not  similar  numbers ;  therefore,  the 
difference  can  only  be  indicated  by  plac- 
ing the  sign,  - ,  between  them. 


EXERCISE  LXXXVIII. 


ibtract : 
1, 

2. 

3. 

4. 

mx  weight 

12b  X  time 

14&Xcost 

105  X  loss 

12  X  weight 

57  X  time 

cost 

49  X  loss 

5. 

6. 

.7. 

8.     • 

9. 

120  X(/ 

224.x  f 

1 75  XamL 

365  Xi; 

459  Xw; 

88  X(/ 

189  X/ 

4&Xamt. 

- 

175X1; 

2m  Xw 

10. 

11. 

12. 

13. 

u. 

U7p 

41 1 

lllr 

215  s 

512  i 

195p 

2St 

55  r 

77  s 

711 

15, 

16. 

17. 

18. 

19. 

75  X  cost 

157  XiVo. 

17 

Xw 

189 1; 

7b  I 

28  X  area 

imx  price             14  Xf 

74.P 

88  7/; 

ADVANCED   ARITHMETIC.  201 

115.   Multiplication. 

EXAMPIiES. 

1.  12  tens 

6  Explanaiion  :    6  x  12  tens  are  72  tens. 

72  tens 

2.  ten                                                           3.    12X cost 
_6  _6 

6  X  ten^  or  6  tens,  product.  72  X  cost,  product. 

j^.    12Xw  5,    12n 

6  6 


72  X  w,  product.  .  72  n,  product. 

6.  Multiply  phj  n. 

Explanation:    (1)  lxp  =  p. 

(2)  2xp  =  2p. 

(3)  3xp  =  3ja;  and  so  on. 

(4)  nxp  =  np,  resuli. 

Thus,  writing  letters  together,  as  nj),  indicates  the  product 
of  n  and  p.  It  is  common  also  to  use  the  expression  nxp  to 
indicate  the  product.    Both  are  correct. 

7.  Multiply  bXp  by  n. 

Result:  nx5x^j  =  5xnx/),  or  5pn. 

Note. — If  p  and  n  are  abstract,  by  Principle  4,  p.  21,  they  may  stand 
in  any  order ;  it  is  customary  to  put  the  numerical  factors  before  the 
letters  or  words. 

8.  Multiply  7c  by  9  a. 

Result  ;9ax7c  =  9xax7xc  =  63 ac. 

Note. — Since  factors  may  stand  in  any  order,  the  7  and  9  may  be  put 
together,  making  63. 


202  ADVANCED   ARITHMETIC. 

EXERCISE  LXXXIX. 


Multiply  : 

1. 

2. 

3. 

4' 

7  X  cost 

24  X  area 

137  tens 

56  X  rate 

12 

86 

48 

17 

•5. 

6. 

7. 

8. 

9. 

mxv 

45Xw 

IbS  Xth 

4&Xl 

129  Xp 

5 

18 

11 

- 

15 

29 

10. 

11. 

12. 

13. 

u. 

bp 

17p 

lb  I 

12pr 

Qlw 

n 

r 
16. 

w 

t 
18. 

h 

15. 

17. 

19. 

pi 

Ih 

6wh 

Qpt 

7wl 

r 

w 

41 

5Xr 

7xh 

11 6,  Division. 

1.  Divide  12  tens  by  6. 

Process  :  61 12  ^ens 

' ,       or,  12  tens +Q  =  ^  tens  =  2  tens,  result. 

2  tens,  result. 

Principle  :  Dividing  any  factor  of  a  continued  'multiplication 
divides  the  product. 

2.  Divide  the  cost  by  6. 

Process  :   Cost  -«-  6  =  — ^— ,  result. 
6 

Note. — The  result  can  only  be  indicated,  and   the  fractional  form  is 
usually  employed. 


ADVANCED   ARITHMETIC.  203 

3.  Divide  5Xc  by  9. 

-,  _  „     5xc         5c  ,, 

Process:  5xc-i-9  =  —Q-,  or, -q,  result. 

4..  Divide  16  ly  by  n. 

Process  :  lQw'+n= ,  result. 

n 

5.  Divide  15^  by  25w;. 

Process  :  15l  +  25w=  -^-  =  ^-  ,  result. 

Note. — Cancel  any  common  factors  found  in  both  numerator  and  de- 
nominator. 


6.  6ln-^7wn=(   )  ? 
Process :  5ln  +  7 

Note. — n  is  common  to  both  numerator  and  denominator. 


Process  :  bin +  7 wn  =  ^ —  =  ,;r— ,  result. 
7  wn    7  w 


EXERCISE  XC. 

■orm  the  process  indicated  : 

1.  64Xco8«-^8. 

11. 

wXl^h. 

2.  96Xpnce-j-12. 

12. 

^Xi^lOXp. 

3.  ^Q>y.value-^Q. 

13. 

i-7-pXr. 

Jf..  20Xweight^6. 

U- 

5x«-^75. 

5.  IXcost-^U. 

15. 

12  Iw-hmih. 

6.   12Xprice-^72. 

16. 

mi^7Xpt. 

7.  cost-^S. 

17. 

ipt-^prt. 

8.   12^^80. 

18. 

12cn^27wn. 

9.  16x^(;-^w. 

19. 

IS  rt -^40  pt. 

10.  w-^l. 

20. 

120  prt-T- 144:  pnt 

117.  Involution. 

EXAMPLES. 

1.  Square  j9. 

Process  :  p  xp  =p^,  result. 
Note. — Read  "p^"  p  square. 

2.  Square  the  number. 

Process  :  Number  x  number  =  (number)^,  result. 


204 


ADVANCED    ARITHMETIC. 


3.  Square  5  a . 

Process  :  5  a  x  5  a  =  25  a^,  result. 


Square  a-\-b. 
Process  : 

a+6 


ab  +  b^ 


a'+ab 
a^-\-2ab  +  b^,  result. 


Questions  :  Do  this  process  and 
result  agree  with  the  principle  on 
page  181  ?  Can  you  repeat  that 
principle  from  memory  ? 


S.  Cube  7  Z. 

Process  :  7 1x71x7 l  =  MSl^,  result. 

Note— Read  ''P''  I  cube. 


6.  Cube 


5a 

w  ' 

„  ba^5a    ba    125  a^  ,^ 

Process :  —  x  —  x  —  =  — ?- ,  result. 

www        w^ 


7. 


7c 
4n 


=  {   ) 


o  7c     7c     7c     7c     7c     168070^ 

Process :  .—  x  r-  x  .—  x ;—  x  ;i—  =  t^^oT-^  >  result. 
4n    4n    4n    4n    4n     1024  n* 


8.   (a+br=(   )? 

Process  : 

a  +  b 

ab-^b^ 
a^-\-ab 


a2+2a6  +  62 
a+b 

a^b  +  2ab^+b^ 
a^+2a%  +  ab^ 

aH3a26+3a62+6^  result. 


Questions  :  Do  this  process  and 
result  agree  with  principle,  page 
182  ?  Can  you  repeat  that  prin- 
ciple from  memory  ? 


ADVANCED    AEITHMETIC. 
EXERCISE  XCI. 


205 


Give  results  orally : 


1.  Square  (1)  ^^y  .(2)   ^n;   (3)  j>n;    (4)   ^;   (5)  |^^. 

2.  Square  (1)  a.^h ;   (2)  r+p;   (8)  n+r;   (4)  t(;+^. 
5.  Cube  (1)    ..;  (2)  5n;    (3)  nl ;    (4)   |^';   (5)  -g. 
4.  Cube  (1)  r/  +  /i;   (2)  r+p;  (3)   n+r;   (4)  w+L 


O^'-ye  luritten  processes  : 

7 


9« 


146 
126^ 


^149  a 

8.    (r/2)2. 

118.  Evolution, 


9. 
10. 
11. 
12. 


4:rl 


9vt 

7c?/7 

Il71j9 


4ar  J 


EXAMPLES. 


i.    -i/a2=(    )  ?  

Process^ :    v  a^  =  V  a  x  «  =  a,  result. 

Question  :  What  is  the  square  root  of  a  number  ? 


2.    V^a^V'={    )  ? 

Process  :  'V^S nhxSab  =  S ab,  result. 


4 


ma' 


>5c2      ^   ^ 

/36a2        /6  a     6  a    6  a  ,^ 

Process  :   ^ =  ^  —  x —  =  —  >  result. 

\25c2     \5c     5  c     5  c 


13.    (2+n)2 
i4.    (3 r/ +4  6)2 

15.  {w-\-6y 

16.  (4r+3-^i)' 


206  ADVANCED    ARITHMETIC. 

4.     VM^={    )  ? 

Process  :    ^Q4a^=  1^4 ax4ax4«  =  4a,  result. 
Question  :  What  is  the  cube  root  of  a  number  ? 


5.  l^^^l^=^ ). 

\216/V     ^   ^  • 


si  27aSr3_  slSar 


T,  o  ,  -.  .V  ,        a  ,„..,     3<7r    Sar    Sar  ,, 

Process  :    ^j =  ^  —  x  —  x  —  =  — ,  result. 

6hv     6  Iw     6  Iw 


6.    -^ SI  wH''={    )  ? 


Process  :  ~y SI wH*  =  'V 3 wl xSwlxSwlxS wl  =  3 wl,  result. 


7.    Va^-i-2ab-\-b^=(   )  ? 

Process  :  V a'^-^2ah-\-h'^  =  -/(a-t-6)(a  +  />)  =  a  +  h,  result. 
Note. — "(a+6)(«  +  6)"  =  a  +  &  multiplied  by  a +  6. 


Process  :  f  a^ +3 a% +3 ab^ -\-h^  =  f  {a -\-h){a-\-h){a-\-h)  =  a  +  h,  result. 

EXERCISE  XCII. 

(T*'?;e  results  orally : 
1.  Square  root  of    (1)    l^ ;    (2)    An^ ;    (8)    10/^;    (4)  ^^  ; 

;^.  Square  root  of  (1)  l'^-\-2lr-\-r'^ ;  (2)    7i2+27ic+c2. 

.?.  Cube   root  of  (1)  r^•  (2)  27 />^•  (8)   125  i/;3/^^•  (4)  ^3; 

^^^  2l6^^* 

^.  Cuberootof  (l)y+3i>V+8pr2+?-3-  (9)  /3+3/2^+3Zw;24_^3^ 


ADVANCED    ARITHMETIC.  207 


Give  written  process  : 


5. 


125  c^]  i 


9.   {p^-}-2rp+r^)^ 


6.    |-^?i^]^  '  10.   (^/;2+4w+4)^ 


7.  ^  11.    {t^^^t^r+Ztr'^+T^)^ 

8.  f_^?_]^  i^.   (71^+3  7iV+3wr2+r3)^ 


119,  Equations  Containing'  Letters. —  Formulas, 
which  are  equations  containing  letters,  are  often  used  in  arith- 
metic.    For  example, 

(1)    C=np. 

Now,  there  are  three  numbers  in  this  equation.  If  any  two 
of  them  are  known,  the  other  may  be  found.     For 

(1)  =  (2)  ■np  =  C. 

(2)^«  =  (8)  p=-^. 

(2)-i>  =  (4)   »  =  y. 

If  the  values  of  n  and  p  be  substituted  (put  in  place  of  the 
letters)  in  (1),  the  result  will  express  the  value  of  C ;  if  the 
values  of  C  and  n  be  substituted  in  (3),  the  result  will  express 
the  value  of  p  ;  and  if  the  values  of  C  and  p  be  substituted  in 
(4),  the  result  will  express  the  value  of  n. 

Note. — When  an  equation,  containing  letters,  is  so  arranged  that  one 
be  expressed  in  terms  of  the  others.  Thus,  in  (1),  C  is  expressed  in  terms 
of  p  and  n;  in  (3),  p  is  expressed  in  terms  of  C  and  n;  and  in  (4),  n  is  ex- 
pressed in  terms  of  C  and  />. 


208  ADVANCED    ARITHMETIC. 

EXAMPLES. 

1.  A=lw.   (1)  Express  the  value  of  w  in  terms  of  A  and  I. 
(2)  Express  the  value  of  I  in  terms  of  A  and  w. 

Process :   (1)  A  =  lw. 
(1)=(2)  lw  =  A. 

(2)  H-  Z  =  (3)  u'  =  A ,  Is t  result. 
(2)  -H  w  =  (4)  Z  =  A ,  2d  result. 

2.  F=—-.     Express  the  values  of  m,  v,  and  t,  each  in  terms 
of  the  other  numbers. 

Process:   (1)  F= — . 

(1)  =  (2)  !^'  =  i^. 
tx(2)  =  (3)  mv  =  Ft. 
(3)+i,'  =  (4)  w  =  -^,  1st  result. 

V 

Ft 
(3)  +  m  =  (5)  V  =  — ,  2d  result. 

(3)  =  (6)  Ft="mv, 

mv 
i6)^F={7)   <  =  -p-,  3d  result. 

S,   432  =  16(2^-1).     Find  the  value  of  «. 
Process  :   (1)  432  =  16  (2 1  - 1). 
(1)  =  (2)  16(2<-1)  =  432. 
(2)-h16  =  (3)  2<-1  =  27. 
Transposing,  we  get  (4)  2t  =  28. 

I-  of  (4)  =  (5)  t  =  14,  result. 

4-  S  =  \gt^.     Express  the  values  of  g  and  t,  each  in  terms  of 

the  other  numbers. 

Process:   (1)  S  =  ^gt\ 
(1)  =  (2)  i,gt''=^S. 

2  Sf 
(2)  +  i<2  =  (3)  ^  =  -_^  Istresult. 

2,S 
(2)-i-f7  =  (4)  <2  =  — 

2^ 


-i/(4)  =  (5)  <  =  ^^,  2d  result 


ADVANCED    ARITHMETIC.  209 


5.  A=p-\-prt.     Express  the  values  of  j9,  i,  and  r. 

Process:     {\)  A=p-\-prt=p  {l-\-rt). 
(i)  =  {2)p(l  +  r0  =  ^. 
(2)^(l+r0  =  (3)  ^^fv;:^'  1st  result. 

{l)  =  {'i)  p+prt  =  A. 

Transposing,  we  get  (5)  prt  =  A-p. 

A  -» 

0)+pr^{Q)  t  = i^,  2d  result. 

pr 

(5)+p<  =  (7)    r  =  —^,  3d  result. 
pt 


EXERCISE  XCIII. 


._  PJ^ 


1.  p—-L~-.      Express  the  values  of /)  and  ^. 

lOU 

2.  F=Ma.     Express  the  values  of  M  and  a. 

3.  W—Fd.     Express  the  values  of  i^and  d. 

4.  W=Mda.     Express  the  values  of  M,  d,  and  a. 

5.  25x30=15  M.     Find  the  value  of  M. 

6.  869.84=1  (/  (24-1).     Find  the  value  of  g. 

7.  1302.48  =  16.08  f.     Find  the  value  of  t. 

W 

8.  A=——-.     Express  the  values  of  IF  and  t. 

9.  V=lwh.     Express  the  values  of  Z,  w,  and  h. 

10.  H^=B^+P\     Express  the  values  of  P  and  B. 

11.  C—~^.     Express  the  values  of  ^  and  i2. 

12.  A  =  irl^.     Express  the  value  of  R. 

Note. — The  Greek  letter  ir  (pi)  is  used  for  the  number  3.1416. 

13.  Find  the  value  of  F  in  No.  8,  if  1^=75  and  d  =  15. 

IJ/..  I=prt.  Find  the  value  of  p,  if  /,  r  and  t  are  respectively 
102,  T«o,  4. 

15.  Find  the  value  of  B  in  No.  10,  if  H  and  P  are  respectively 
50  and  80.      . 


210  ADVANCED    ARITHMETIC. 

II.    STUDY  OF  PROBLEMS. 
A.     DEVELOPMENT  AND  USE  OF  FORMULAS. 

120.  General  Numbers  and  General  Prob- 
lems.—  If  I  say,  "  The  fence  is  1^.0  rods  long,^^  I  use  a,  jjarticular 
number  J  J^O  rods.  If  I  say,  "  The  fence  is  I  rods  long,"  I  use  a 
general  number,  I  rods,  which  may  prove  to  be  40  rods,  or  some 
other  number  of  rods.  This  general  form  of  expressing  a  num- 
ber that  may  vary  in  its  numerical  value  is  of  great  service  in 
mathematics. 

Examine  the  following  problems : 

1.  What  will  5  books  cost,  at  $2  each  ? 

2.  What  will  25  lb.  of  meat  cost,  at  10/  per  lb.? 

3.  Find  the  value  of  140  A.  of  land,  at  $40  per  A. 
4..  What  will  12  horses  sell  for,  at  $75  each  ? 

These  are  all  different  problems,  yet  they  all  belong  to  the 
same  type. 

Type  :  What  will  a  given  number  of  articles  cost,  at  a  given  price 
each  ? 

By  using  letters  to  represent  the  numbers  in  this  type,  it  may 
be  expressed  in  the  following  form : 

What  will  n  things  cost,  at  $p  {or  pf^)  each  f 

Such  a  type,  which  is  a  representative  of  a  great  class  of 
problems,  is  often  spoken  of  as  a  General  Problem.  It  is  of 
great  service,  in  solving  particular  problems,  to  know  to  what 
type  they  belong. 

EXAMPIiES. 

1.  What  is  the  area  of  a  field  50  rd.  long,  80  rd.  wide  ?  Give 
the  type. 

Type  :  What  is  the  area  of  a  surface  (or  field)  I  rd.  long,  w  rd.  wide? 

Note. — In  giving  types,  it  is  well  to  use  initial  letters,  as  far  as  con- 
venient, to  represent  the  numbers ;  as,  I  for  length  and  w  for  width. 


ADVANCED    ARITHMETIC.  211 

2.  Find  the  interest  on  $400  for  2^  years  at  7%  per  annum  ? 
Give  the  type. 

Type  :   Find  the  interest  on  %p  for  t  yr.  at  r  %  per  annum. 

3.  Find  the  length  of  a  solid  of  8600  cu.  ft.,  20  ft.  wide,  6  ft. 
thick. 

Type:  Find  the  length  of  a  solid  of  v  cu.  ft.,  w  ft.  wide,  t  ft.  thick. 

EXERCISE  XCIV. 

Give  the  types : 

1.  Find  20%  of  $500. 

2.  How  many  books  will  bring  $40  at  $5  each  ? 

3.  $75  is  15%  of  what  number  ? 

4.  What  must  be  the  width  of  one  acre  (160  sq.  rd.),  if  it  is 
40  rd.  long  ? 

5.  What  force  is  required  to  add  10  ft.  per  second  to  the  ve- 
locity of  15  lb.  of  matter  ? 

Note. — Use  a  (acceleration)  for  the  number  of  feet,  and  J/ (mass)  for 
the  number  of  pounds. 

6.  75  is  what  %  of  150  ? 

7.  I  sold  20  lb.  of  butter  for  $2.     Find  the  price  per  lb. 

8.  Find  the  width  of  a  wall  of  2340  cu.  ft.,  130  ft.  long,  1^ 
ft.  thick. 

121.   Solving  tlie  General  Problem  or  Type. 

The  same  reasoning  and  plan  are  used  in  solving  the  general 
problem^  or  type^  as  are  used  in  solving  thQ  particular  problem. 

EXAMPLE. 

Find  the  cost  of  n  things,  at  $p  each. 

Solution:    (1)  Cost  of  n  things,  at  $p  each  =  $(  )?    (Question.) 
(2)  Cost  of  1  thing,  at  $1  each  =  $1.    (Basis.) 
px{2)  =  (3)  Cost  of  1  thing,  at  '^p  each  =  $p. 
nx(3)  =  (4)  Cost  of  n  things,  at  $p  each  =  $np,  result. 


212  ADVANCED    ARITHMETIC. 

If  the  cost  be  represented  by  C  and  (4)  be  expressed  in  its 
abstract  form,  it  becomes  — 

(5)   C=np. 

This  equation  is  called  a  rormula,  and  it  expresses  an  ab- 
stract numerical  relation  that  is  always  true. 

Relation  :  The  cost  of  a  number  of  things  of  the  same  kind  is 
equal  to  the  product  of  the  number  of  things  and  the  price  of  one  of 
the  things. 

This  formula  relates  three  numbers :  the  cost,  the  number  of 
things,  and  the  price  of  each.  When  any  two  of  these  numbers 
are  given,  the  third  may  be  found  by  aid  of  the  formula. 

EXAMFIjES. 

1.  Find  the  cost  of  20  bu.  of  apples,  at  30/  per  bu. 

Solution:     (1)  C  =  np. 

(2)  0  =  20x30  =  600. 

.-.  the  cost  is  600^,  or  $6. 

2.  How  many  bushels  of  apples  at  $.30  each  can  I  buy  for 

$18? 

Solution:    (!)  C-np. 
(2)  18=.3n. 
(2)  =  (3)  .3ri  =  18. 

(3)-H.3  =  (4)  n  =  ~  =  60. 

.'.  the  reqd.  answer  is  60  ba. 

3.  If  50  bu.  of  apples  cost  $15,  find  the  price  per  bushel. 

Solution:     (1)  C  =  np. 
(2)  15  =  50p. 
(2)  =  (3)  50j9  =  15. 
5Vof(3)  =  (4)/>  =  B  =  .3. 

.'.  the  price  is  $.3,  or  30^. 

Note. — In  examples  2  and  3,  the  work  would  be  shortened  by  one 
equation  if  the  formula  had  been  turned  around  in  the  J3rst  equation  so 
that  the  required  number  would  appear  in  the  left  member. 


ADVANCED    ARITHMETIC. 


213 


EXEEOISE  XCV. 

1.  If  n  bu.  of  apples  cost  $C,  find  the  price  per  bu.     Answer: 
C 

^     n  - 

2.  If  $C  are  paid  fdr  apples  at  $p  per  bu.,  find  the  number 

n 

of  bushels  bought.     Answer:  n=  — . 

Use  answer  to  No.  1  as  a  formula. 

3.  Find  the  price  per  acre,  if  120  A.  of  land  sell  for  $1440. 
J^.  I  sold  5000  lb.  of  bacon  for  $400.     Find  the  price  per  lb. 

Use  answer  to  No.  2  as  a  formula  : 

5.  I  spend  $15.60  for  turkeys,  at  $.65  each.     How  many  did 
I  buy  ? 

6.  A  dealer  invests  $975  in  buggies,  at  $75  each.    How  many 
did  he  buy  ? 

B.  MENSURATION. 

1.  LINES  AND  ANGIiES. 


122.   Definitions. — Define  a  line.     (See  p. 149.) 


A  line  whose  length  has  but  one  direction 
is  a  Straiglit  Line. 

A  line  whose  length  changes  direction  at 
every  point  is  a  Curved  Line. 


AB.,  a  line; 
also,  a  straight  line. 


CD,  a  curved  line. 


Two  lines  that  never  approach  each  other 
are  Parallel. 

Note. — Lines  that  are  parallel  have  the  same  di- 
rection and  never  meet.  If  two  lines  meet,  they 
form  one  or  more  angles,  and  have  different  di- 
rections. 

An  Angle  is  the  difference  of  direction 
between  two  lines  that  meet. 


AB  and  CD,  parallel 
lines. 


AC  By  an  angle. 


214 


ADVANCED   ARITHMETIC. 


Note  1. — As  line  BC,  starting 
from  the  position  AB,  swings 
around  point  C,  the  direction  of 
line  BC  is  continually  changing, 
and  the  difference  of  direction  be- 
tween line  AC  and  line  BC  is  con- 
tinually growing  larger.  Angle 
^05  is  smaller  than  angle  ACB^ ; 
angle  ACB^  is  smaller  than  angle 
ACB^^;  and  so  on. 

Note  2.— If  two  lines,  coming  together,  stop  at  the  point  of  meeting, 
they  form  but  one  angle.  If  only  one  line  stops  at  the  point  of  meeting, 
they  form  two  angles.  If  neither  line  stops  at  the  point  of  meeting,  they 
form  four  angles. 


One  angle. 


Two  angles. 

Four  angles. 

When  the  directions  of  two  lines  that  meet  are  such  that,  if 
the  lines  are  continued  past  the  point  of  meeting,  they  form 
equal  angles,  the  lines  are  Perpendicular  to  each  other. 
When  the  angles  thus  formed  are  not  all  equal,  the  lines  are 
Oblique  to  each  other. 

B 


AC  and  BC  are  perpendicu- 
lar to  each  other. 


AC  and  BC  are  oblique  to 
each  other. 


Note.  —  The  lines  forming  an  angle  are  called  its  sides,  and  the  point 
of  meeting  of  the  sides  is  called  the  vertex  of  the  angle. 


ADVANCED    ARITHMETIC, 


215 


ACB^  a  right  angle. 


B 


I 


/ 


ACB^  an  acute  angle. 


ACB,  an  obtuse  angle. 


A  Riglit  Angle  has  its  sides  perpendicular  to  each  other. 
An  Acute  Angle  is  smaller  than  a  right  angle. 
An  Obtuse  Angle  is  larger  than  a  right  angle. 

Note.— The  common  unit  of  angular  measure  is  the  degree,  marked  (°). 
1  right  angle  =  90° 

A  line  parallel  with  the  regular 
lines  of  writing  or  print  is  a  Hori- 
zontal Line.  . „ 


A  line  perpendicular  to  the  regular 
lines  of  writing  or  print  is  a  Ver- 
tical Line. 


AB,  a  horizon- 
tal line. 


CD,  a  vertical  line. 


Note.— The  word  "  horizontal "  came  to  be  applied  to  lines,  probably 
because  in  writing  upon  upright  surfaces  such  as  blackboards,  such  lines 
are  parallel  with  the  horizon ;  but  if  the  surface  on  which  you  write  is 
parallel  with  the  horison,  all  lines  made  on  that  surface  are  also  parallel 
with  the  horizon. 

EXERCISE  XOVT. 


Define  and  draw  (free-hand  )  : 

1.  A  line. 

2.  A  straight  line. 

3.  A  curved  line. 

4..  Two  parallel  lines. 

5.  An  angle. 

6.  Two  perpendicular  lines. 


7.  Two  oblique  lines. 

8.  A  right  angle. 

9.  An  acute  angle. 

10.  An  oblique  angle. 

11.  A  horizontal  line. 

12.  A  vertical  line. 


Note.— Become  familiar  with  these  lines  and  angles  before  passing  to 
the  next  subject. 


216 


ADVANCED    ARITHMETIC. 


2.  PLANE  SURFACES. 


123.    Parallelogram.— Define  Surface.     (See  p.  154.) 

A  Plane  is  a  surface  each  of  whose  dimensions  has  but  one 
direction. 

A  Polygon  is  a  plane  bounded  by  straight  lines.  The 
straight  lines  are  called  sides,  and  the  points  where  the  sides 
meet  are  called  angles  or  vertices. 

^ ^C 

A  Quadrilateral  is  a  polygon  of 
four  sides.  The  lines  joining  the  al- 
ternate angles  are  called  diagonals. 

ABCD,  a  quadrilateral ;  AC  a 
diagonal. 

A  Parallelogram  is  a  quadrilateral  whose  opposite 
sides  are  parallel.  The  line  on  which  a  ^parallelogram  rests  is 
usually  called  its  base,  and  the  perpendicular  distance  from 
the  base  to  the  opposite  side  (or  opposite  side  produced)  is 
called  its  altitude. 

A  Rectangle  is  a  parallelogram  whose  angles  are  right 
angles. 

A  Square  is  a  rectangle  whose  sides  are  equal. 
DEC  L  C  ^ ^ 


B 


ABCD,   a    parallelogram    and 

rhomboid ;  AB,  the  base  ; 

EF,  the  altitude. 


ABCD,  a  rectangle ;  AB,  the 
base ;  DA  or  CB,  the  al- 
titude. 


B 

ABCD,  a  square  ;  AB,  base  | 
CB  or  AD,  the  altitude. 


A  Rhomboid  is  a  parallelogram  whose 
angles  are  not  right  angles. 


For  form  of  rhomboid, 
see  parallelogram. 


ADVANCED    ARITHMETIC. 


A  Rhombus  is  a  rhomboid  whose 
sides  are  equal. 


ABCDy  a  rhombus;  AB,  the 
base ;    DE,  the  altitude. 


EXERCISE  XCVII. 

Define  and  draw  (free-hand,  hut  carefully)  : 


1.  A  quadrilateral. 

2.  A  parallelogram. 

3.  A  rectangle. 

Define: 

7.  A  surface. 

8.  A  plane. 

9.  A  side. 
10.  A  vertex. 


4.  A  square. 
6.  A  rhomboid. 
6.  A  rhombus. 


11.  A  diagonal. 

12.  A  polygon. 

13.  A  base. 
IJ^.  An  altitude. 

Note. — These  terms  must  be  defined  as  applied  to  polygons. 

Problem  1 :  Letting  b  represent  the  length  and  a  the  width  of  a 
rectangle,  develop  the  formula  for  the  area. 

BY   EQUATIONS. 

Solution:  (1)  Area  of  a  surface  h  ft.  1.,  a  ft.  w.  =  (  )  sq.  ft.? 

(Question). 
(2)  Area  of  a  surface   1   ft.   1.,   1  ft.  w.  =  l  sq.  ft. 
(Basis). 
6x(2)  =  (3)  Area  of  a  surface  h  ft.  1.,  1  ft.  w.  =  6  sq.  ft. 
ax(3)  =  (4)  Area  of  a  surface  h  ft.  1.,  a  ft.  w.  =a6  sq.  ft., 
answer. 

BY   PROPORTION. 

Solution :  (1)  Area  of  a  surface  6  ft.  1.,  a  ft.  w  =  ( )  sq.  ft.?   (Question.) 

(2)  Area  of  a  surface  1  ft.  1.,  1  ft.  w.  =  1  sq.  ft.     (Basis.) 

(3)  6xa:lxl::(  ):  1? 

(4)  ^^^^^  =  6a,  answer. 
^  '      1x1 

Using  A  for  area,  the  result  may  be  written  as  follows : 

A  =  ba. 


if  UWiVERSlTV 


218  ADVANCED    ARITHMETIC. 

Relation  I.  Abstractly,  the  area  of  a  rectangle  is  equal  to  the 
product  of  the  base  and  altitude. 

EXAMPLES. 

1.  Find  the  area  of  a  wall  20  ft.  long  and  7^  ft.  high. 
Solution:   (1)  A  =  ba. 

(2)  ^  =  20x7H150. 
.'.  the  required  area  is  150  sq.  ft. 

2.  A  rectangular  field  is  80  rd.  long  and  contains  10  A.  How 
wide  is  it? 

Solution :  (1)  10  A  =  1600  sq.  rd. 

(2)  ab  =  A. 

(3)  80  a  =  1600. 
(3)  +  80  =  (4)  a  =  20. 

.'.  the  required  width  is  20  rd. 

3.  A  sidewalk  4  ft.  wide  contains  960  square  feet.  How 
many  yards  long  is  it? 

Solution:  (1)  ab  =  A. 
(2)  4  6  =  960. 
iof  (2)  =  (3)  6  =  240. 

.'.  the  required  length  is  240  ft.  or  80  yd. 

Note. — Before  applying  the  formula,  the  dimensions  must  be  ex- 
pressed in  linear  units  of  the  same  denomination;  then,  the  area  will  be 
square  units  of  that  denomination.  Use  only  the  corresponding  abstract 
numbers  in  the  formula. 

Problem  2 :  Develop  the  formula  for  the  square. 

Development :  A  square  is  a  rectangle,  whose  base  (6)  and  altitude  (a) 
are  equal.  If  we  call  one  side  of  the  square  s,  we  can  substitute  s  for  a 
and  b  in  the  formula,  thus : 

A  =  sxs;  or, 
A  =  sK 

Relation  II.  Abstractly,  the  area  of  a  square  equals  the  square 
of  one  of  its  sid^s. 


ADVANCED    ARITHMETIC.  219 


EXAMPLES. 


1.  Find  the  area  of  a  square  one  of  whose  sides  is  46  ft. 

Formula :  (1)  A=^s^. 

(2)^  =  462  =  2116. 

.',  the  required  area  is  2116  sq.  ft. 

2.  The  area  of  a  square  is  102400  sq.  rd.     Find  the  length  of 

one  side. 

Formula:  (1)  s^  =  A. 

__    (2)  82=102400. 
-i/(2)  =  (3)  s  =  320. 

.  •.  the  required  result  is  320  rd. 

Problem  3:  Develop  the  formula  for  the  parallelogram. 

Development :  Let  A  BCD  be  a  par-      F                D              F  C 

allelogr am,  whose  base  is  ^-B.  Draw     r '/ j ~7 

BE  from   B  perpendicular  to   CD,     \  y/^  \  ^ 

cutting  off  the  part  BEC.    Place  this     !     y/^  \       >^ 

part  on  the  left  as  shown  in  AFD,     \y^ I/^ 

forming  the  rectangle  J  J5^i^.   Thus,     /|  5 

the  parallelogram  has  been  changed 

to  a  rectangle,  whose  hase  and  altitude  are  the  same  as  those  of  the  par- 
allelogram. Then,  the  area  of  the  parallelogram  is  equal  to  that  of  the 
rectangle;  or, 

A=ba. 

Relation  III.     The  same  as  that  of  the  formula  for  a  rectangle. 

EXAMPLES. 

1.  Find  the  area  of  a  parallelogram  whose  base  and  altitude 
are  25  ft.  and  50  ft.  respectively. 

Formula:    (1)  A  =  ba. 

(2)  ^  =  25x50=1250. 

.'.  the  reqd.  area  is  1250  sq.  ft. 

2.  The  area  of  a  parallelogram  is  1440  sq.  rd.     What  is  its 

length,  if  it  is  only  16  rd.  wide  ? 

Solution  :     (1)  16  &  =  1440.     ( Why  ?) 
(2)  b  =  m^  =  90. 
.•.  the  reqd.  length  is  90  rd. 

Note.— In  rapid  work,  the  formula  may  be  omitted,  as  in  example  2; 
but  never  until  its  use  is  thoroughly  understood — not  until  the  pupil 
knows  just  what  substituting  in  a  formula  means. 


220  ADVANCED    ARITHMETIC. 

EXERCISE  XOVIII. 

1.  In  what  respect  is  a  square  a  parallelogram  ?  In  what 
respect  a  rectangle  ? 

2.  What  is  the  difference  between  a  square  and  a  rhombus  ? 
S.  Is  it  true  that  the  formula  A=ah,  will  apply  to  a  rectan- 
gle ?     A  square  ?     A  rhomboid  ?     A  rhombus  ?     Why  ? 

4-.  Show  how  the  formula  A=ah  is  modified  to  ^=8^  for  the 
square. 

5.  Carefully  draw  and  cut  out  of  paper  a  parallelogram ;  then, 
cut  it  into  two  parts  and  put  the  parts  together  to  form  a  rec- 
tangle as  is  indicated  in  problem  3.  Can  every  parallelogram 
(not  already  a  rectangle)  be  thus  reduced  to  a  rectangle  ? 

6.  In  figure  in  problem  8,  does  the  perpendicular  BE  have  to 
be  erected  at  B,  or  may  it  be  erected  somewhere  else  along  the 
base  ? 

7.  Commit  the  relations  I  to  III. 

Find  the  area  of  the  following  parallelograms  : 

8.  Base  90  ft.,  altitude  37  ft. 

9.  Base  3  ft.,  altitude  12  in. 

Note. — Dimensions  must  be  expressed  in  the  same  denomination  be- 
fore applying  the  formula. 

10.  Base  5  mi.,  altitude  700  rd. 

11.  Base  5i  yd.,  altitude  16^  ft. 

In  each  of  the  following  parallelograms ^  find  the  dimension  not 


12.  Area  2700  sq.  m.,  altitude  60  m. 

13.  Area  960  A.,  base  600  rd. 
IJ^.  Area  1156  sq.  ft.,  base  34  ft. 

15.  Area  100500  sq.  yd.,  altitude  450  ft. 

16.  Area  57500  sq.  cm.,  base  46  dm. 


ADVANCED    ARITHMETIC. 


221 


124.  Triangles.— A  Triangle 

is  a  polygon  of  three  sides.  Any  side 
of  a  triangle  may  be  considered  its  base, 
and  the  perpendicular  from  any  vertex 
to  the  opposite  base  is  the  altitude  cor- 
responding to  that  base. 


ABCf  a  triangle ;   AB,  the  base  ; 
C£,  the  altitude. 


ABC,  an  equilateral  trian- 
gle ;  also,  equiangular. 


ABC,  an  isosceles  tri- 
angle. 


ABC,  a  scalene  triangle;  AB, 
the  base  :  CE,  the  altitude. 


An  Equilateral  Triangle  has  its  three  sides  equal. 
An  Isosceles  Triangle  has  two  of  its  sides  equal. 
A  Scalene  Triangle  has  no  two  of  its  sides  equal. 

Every  triangle  has  three  angles,  which  together  measure  just 
180°. 

An  Equiangular  Triangle  has  its  three  angles  equal. 

(  For  form  of  equiangular  triangle,  see  equilateral  triangle. ) 

NoTB.—  Each  angle  of  an  equiangular  triangle  measures  just  60°. 


A  Riglit  Triangle  has  one  of  its 

angles  a  right  angle.  The  side  opposite 
the  right  angle  is  called  the  hyj}otenuse  ; 
the  other  two  sides  are  usually  called 
base  and  perpendicular. 


ABC,  a  right  triangle;  AB, 
the  base ;  CA,  the  perpendicu- 
lar or  altitude ;  CB,  the  hypote- 
nuse. 


222 


ADVANCED    ARITHMETIC. 


ABC,  an  obtuse  triangle.  ABC,  an  acute  triangle. 

An  Obtuse   Triangle   has  one  of  its  angles  an   obtuse 
angle. 

An  Acute  Triangle  has  all  of  its  angles  acute  angles. 


EXERCISE  XOIX 
Define  and  draw  : 

1.  A  triangle. 

2.  An  equilateral  triangle. 

3.  An  isosceles  triangle. 
4-.  A  scalene  triangle. 


5.  An  equiangular  triangle. 

6.  A  right  triangle. 

7.  An  obtuse  triangle. 

8.  An  acute  triangle. 


9.  Define    (1)    base,   (2)    altitude,   (8)   vertex,   (4)    hypote- 
nuse, (5)  perpendicular. 

10.  How  many  degrees  in  all  the  angles  of  a  triangle  ? 

11.  Can  two  of  the  angles  of  a  triangle  be  right  angles  ?     Can 
two  of  them  be  obtuse  ?     Why  ? 

12.  Construct  an  obtuse  triangle  and  draw  its  three  altitudes. 

Problem  1 :  Develop  the  formula  for  the  triangle. 

Development :  Let  ABC  be  a  triangle. 
Upon  the  base  AB  construct  the  rectangle 
ABDE,  just  as  high  as  the  triangle.  From  C 
drop  a  perpendicular  CF  to  the  base  AB. 
Parts  1  and  2  are  equal,  and  parts  3  and  4  are 
equal.  Therefore,  the  triangle  is  just  half  as 
large  as  the  rectangle.    But  the  base  and  al-  ^ 

titude  of  the  triangle  are  the  same  as  those  of  the  rectangle.    Therefore, 

,      ah 
A  =  -^. 


ADVANCED    ARITHMETIC.  223 

Relation  IV.  Abstractly,  the  area  of  a  triangle  is  equal  to  the 
product  of  its  base  and  altitude,  divided  by  2. 

EXAMPLES. 

1.  Find  the  area  of  a  triangle  whose  base  and  altitude  are 
respectively  40  ft.  and  15  ft. 

Solution:   ^  =  ^^^^^  =  300. 
2 

.'.  the  required  area  is  300  sq.  ft. 

2.  The  base  of  a  triangle  is  75  ft.,  its  area  is  400  sq.  yd. 
Find  its  altitude. 

Solution  :   (1)  400  sq.  yd.  =3600  sq.  ft. 

(2)  1^  =  3600. 

(3)  q  =  36(X)x2^Qg 

.*.  the  required  altitude  is  96  ft. 

Relation  V.  In  right  triangles,  abstractly,  the  square  of  the 
hypotenuse  equals  the  square  of  the  base  plus  the  square  of  the  per- 
pendicular. 

Note.— The  development  of  this  relation  is  too  difficult  to  be  given 
before  the  pupil  has  studied  geometry .    It  gives  us  an  important  formula : 

EXAMPLES. 

1.  The  base  of  a  right  triangle  is  86  ft. ;  the  altitude,  27  ft. 
Find  the  hypotenuse. 

Solution:   (!)  H^  =  B^-\-P^. 

(2)  iy2  =  362+ 272  =  2025. 

(3)  if=  v^2025  =  45. 

.'.  the  required  result  is  45  ft. 

2.  Find  the  diagonal  of  a  square,  one  of  whose  sides  is  32 
inches. 


224  ADVANCED    ARITHMETIC. 


Note. — ABCD  is  a  square.  AC,  the  diagonal,  is 
the  hypotenuse  of  the  right  triangle  CAB.  AB  =  S2, 
BC  =  B2. 

Solution:   (1)  if 2  =  322+322  =  2048. 
(2)  iy=V2048  =  45.25  +  . 
.*.  the  diagonal  is  45.25  in. 

3.  Find  the  base  of  a  right  triangle  whose  perpendicular  is 
52  rd.  and  whose  hypotenuse  is  65  rd. 

Solution:   (1)  B^+T^  =  H^. 

(2)  ^2+522  =  652. 

(3)  ^2  =  652  _  522  =1521. 

(4)  B  =  Vlb21  =  S9. 

.'.  the  required  result  is  39  rd. 

4-.  The  diagonal  of  a  square  is  60  yd.     Find  one  side. 

Note.— The  diagonal  is  the  hypotenuse  of  the  right  triangle  whose 
sides  are  equal. 

Solution:   (1)  8^+8^  =  H^;  or, 

(2)  2s2  =  iy2  =  602  =  3600. 
iof  (2)  =  (3)  82  =  1800. 

(4)  s=Vl800  =  42.426  +  . 

.'.  the  reqd.  side  is  42.426  yd. 

EXERCISE  0. 

1.  Study  the  formula  for  the  area  of  a  triangle  till  you  can 
develop  it  from  memory. 

2.  Find  the  base  of  a  triangle  whose  area  is  5700  sq.  ft.  and 
an  altitude  of  76  ft. 

3.  Find  the  altitude  of  a  triangle  whose  base  is  40  rd.  and 
whose  area  is  5  A. 

4..  Find  the  perpendicular  of  a  right  triangle  when  the  base 
and  hypotenuse  are  respectively  80  ft.  and  50  ft. 

5.  Find  the  base  when  the  perpendicular  and  hypotenuse  are 
respectively  45  ft.  and  90  ft. 


ADVANCED    ARITHMETIC. 


225 


6.  Find  the  hypotenuse  when  the  base  and  perpendicular  are 
respectively  25  in.  and  85  in. 

7.  Find  the  diagonal  of  a  rectangle  whose  base  and  altitude 
are  respectively  10  ft.  and  86  ft. 

8.  Find  the  altitude  of  a  rectangle  whose  base  and  diagonal 
are  respectively  87  ft.  and  90  ft. 

9.  A  house  measures  20  ft.  from  north  to  south ;  15  ft.  from 
east  to  west,  and  12  ft.  high.  Find  the  distance  on  the  floor 
from  the  northwest  to  the  southwest  corner. 

10.  In  the  house  described  in  No.  9,  find  the  distance  from 
the  northwest  lower  corner  to  the  southeast  upper  corner. 

11.  Commit  to  memory  Relations  IV  and  V. 

125.  The  Other  Quadrilaterals. — A  Trapezoid 

is  a  quadrilateral,  having  only  two  parallel  sides.     The  parallel 
sides  are  called  bases. 

A  Trapezium  is  a  quadrilateral,  having  no  two  of  its  sides 
parallel. 

Problem:  Develop  the  formula  for  the  trapezoid. 


Development :  Let  A  BCD  be  a  trap- 
ezoid ;  AB,  lower  base  (B) ;  DC,  upper 
base  (b);  and  FE,  altitude.  The  di- 
agonal BD  divides  the  trapezoid  into 
two  triangles.  In  the  triangle  DAB, 
having  for  its  base  AB  {B)  and  its  al- 
titude FE  (a),  we  have — 


Area  = 


aB 


In  the  triangle  BCD,  having  for  its  base  DC  (b)  and  for  its  altitude  FE 
(a),  we  have — 

.  ab 

Area  =  — . 
2 


226  ADVANCED   ARITHMETIC. 

But  the  sum  of  the  areas  of  the  two  triangles  is  the  area  of  the  trape- 
zoid.    Then, 

,  _aB    ah _aB+ah 

Factoring  a  out  of  both  terms  of  the  numerator  of  the  fraction,  we  have 
the  formula : 

.      a{B-\-h) 


Relation  VI.  '  Abstractly y  the  area  of  a  trapezoid  is  equal  to  the 
product  of  the  altitude  and  the  sum  of  the  bases,  divided  by  2. 

There  is  no  formula  for  a  trapezium.  If  one  of  its  diagonals 
and  the  altitudes  from  the  other  vertices  upon  that  diagonal 
are  known,  its  area  may  be  easily  found  by  the  formula  for  tri- 
angles. 

EXAMPLiES. 

1.  Find  the  area  of  a  trapezoid,  if  its  altitude  is  26  ft.  and 
its  bases  are  50  ft.  and  75  ft. 

Solution  :  A  =  ^^  ^^"^^^^  =  1625. 
2 

.*.  the  reqd.  area  is  1625  sq.  ft. 

2.  Find  the  upper  base  of  a  trapezoid  whose  lower  base  is  35 
in.,  altitude  15  in.,  and  area  860  sq.  in. 

Solution:  (1)  i5i^±^  =  360. 

2x(l)  =  (2)  15  (35 +  6)  =  720.. 
T^sof  (2)  =  (3)  35  +  6  =  48. 

(4)  6  =  48-35  =  13. 

.•.  the  reqd.  base  is  13  in. 

3.  Find  the  area  of  a  trapezium  whose  diagonal  is  48  ft.  and 
the  altitudes  upon  that  diagonal  are  respectively  10  ft.  and  17 
ft. 


ADVANCED   ARITHMETIC. 


227 


Explanation  :  Let  ABCD  be  the  trapezium  ;  DB,  the  diag- 
onal ;  and  BA  and  CF,  altitudes.  In  the  triangle  DBC,  we 
have  given  the  base  1)5,  and  altitude  CF;  and  in  the  triangle 
DAB,  we  have  the  base  DB,  and  altitude  EA. 


SOLUTION. 


.     48x17 
1st  triangle  :  A  = =  408. 

2d  triangle  :  A  = =  240. 


. '.  area  of  trapezium  =  408  sq.  ft.  +240  sq.  ft  = 
648  sq.  ft. 

EXERCISE  CI. 

1.  Define  and  draw  a  trapezoid ;  draw  one  diagonal  and  the 
altitude. 

2.  Define  and  draw  a  trapezium. 

3.  Find  the  area  of  a  trapezoid  whose  bases  are  46  yd.  and  80 
yd.,  and  whose  altitude  is  86  yd. 

4..  Find  the  altitude  of  a  trapezoid  whose  bases  are  30  ft.  and 
50  ft.,  and  whose  area  is  480  sq.  ft. 

5.  Find  unknown  base  of  a  trapezoid  whose  altitude  is  24 
in.,  area  912  sq.  in.,  and  given  base  45  in. 

6.  Find  the  area  of  a  trapezium  whose  diagonal  is  78  ft.,  and 
altitudes  upon  that  diagonal  are  43  ft.  and  21  ft. 

7.  Study  until  you  can  solve  from  memory  the  problems  of 
this  Article. 

126.  Regular  Polygons  and  Circles. — A  Reg- 
ular Polygon  has  its  sides  equal 
and  its  angles  equal.  A  regular  poly- 
gon may  always  be  separated  into 
equal  isosceles  triangles  by  drawing 
lines  from  its  center  to  each  vertex. 
A  line  from  the  center,  perpendicular 
to  a  side,  is  called  the  apothem.  The 
lines  bounding  a  polygon  are  called  its 
perimeter. 

Note. —  A  pentagon   has  5  sides;   a  hexagon,  6  sides 
sides ;  an  octagon,  8  sides ;  a  decagon,  10  sides. 


a  heptagon, 7 


228 


ADVANCED    ARITHMETIC. 


A  Circle  is  a  plane  bounded  by  a  curved  line,  all  parts  of 
which  are  equally  distant  from  a  point 
in  the  plane.  The  point  is  called  the 
center  and  the  curved  line  bounding  the 
circle  is  called  the  circumference.  A 
straight  line,  passing  through  the  center 
of  a  circle  and  terminating  at  each  end 
in  the  circumference,  is  called  a  diam- 
eter. A  straight  line  extending  from 
the  center  to  the  circumference  is  called 
a  radius. 


ADBC,  a  circumference  ;  AB, 
diameter  ;  OC,  a  radius. 


Peoblem  1 :  Develop  the  formula  for  the  regular  polygon. 


Development :  Let  ABCDFG  be  a  regular 
polygon.  Divide  it  into  equal  triangles  by 
drawing  lines  from  the  center  to  each  vertex. 
Draw  OE,  the  apothem,  which  is  the  length 
of  the  altitude  of  each  triangle.  Represent 
the  bases  of  the  triangles  by  6,  b^,  h'^ ^  etc. 
Then, 


In  1st  triangle  :  A  =  -^ 

ah' 
In  2d  triangle  :  A  =  -jr- 

In  3d  triangle  :  A  =  — — 


^Q 


,  and  so  on. 


The  whole  area  = 


ah+a¥  +  a¥'-[-etc 


;  or, 


a{h  +  ¥^¥'^etc.) 


But  h^-y-\-¥'  +  etc.  =  the  perimeter  (P)  of  the  polygon.    Then,  the  for- 
mula for  the  polygon  is  — 

aP 
2  • 


A=- 


Relation  VII.     Abstractly^  the   area   of  a  regular  polygon  is 
equal  to  the  product  of  its  apothem,  and  perimeter,  divided  by  2. 


ADVANCED    ARITHMETIC. 


229 


Relation  VIII.      The  ratio  of  the  circumference  (C)  of  a  circle 
to  its  diameter  (D)  is  S.l^ld  (t). 

Note  1. —  The  character  ir  is  the  Greek  letter  pi,  and  is  used  for  3.1416, 
in  order  to  shorten  the  work  in  writing  formulas. 

Note  2. —  The  development  of  this  relation  is  too  difficult  to  be  given 
here.    It  is  found  in  geometry. 


Formula  : 


IT  =-^=8.1416. 


Since  the  diameter  equals  two  times  the  radius  (R),  we  may 
substitute  2  R  for  D. 

C 

C=2^R. 

Relation  IX.      The  circumference  of  a  circle  is  equal  to  2  times 
SAJf-lG  times  the  radius. 

Problem  2:    Develop  the  formula  for  the  area  of  the  circle. 


Development :  The  figures  above  suggest  the  idea,  that  as  the  number 
of  sides  of  a  regular  polygon  increase,  the  polygon  approaches  the  form 
of  a  circle.  Proceeding  upon  the  suggestion,  and  considering  a  circle  a 
regular  polygon  of  an  infinite  number  of  sides,  we  observe  that  the 
apothem  becomes  the  radius  of  the  circle  and  the  perimeter  becomes  the 
circumference  and  the  formula  for  the  polygon. 

aP 

^: 

2  ' 


A  = 


becomes  A  = 


230  ADVANCED    ARITHMETIC. 

But  from  the  last  formula  we  know  that  the  circumference  equals 
2  TT  /?.     Putting  this  in  the  place  of  C,  we  have-^ 

A  = ;  or, 

2       ' 

Relation  X.  Abstractly j  the  area  of  a  circle  is  equal  to  3.14.16 
times  the  square  of  the  radius. 

EXAMPIiEB. 

1.  Find  the  area  of  a  regular  pentagon  whose  apothem  is  10 
ft.  and  one  side  14.58  ft. 

Solution  :  (1)  P= 5  x  14.53  ft.  =  72.65  ft. 

(2)  ^  =  1«^  =  363.25. 

.'.  the  area  is  363.25  sq.  ft. 

2.  Find  the  diameter  of  a  circle  whose  circumference  is 
219.912  rods. 

Solution:   (1)  -^-  =  3.1416. 

-     (2)   219.912  ^3  .^^^g 

(3)  219.912  =  3.1416  1). 

(4)  2)=  219.912  ^^Q 
'  3.1416 

^  .'.  the  diameter  is  70  rd. 

S.    Find  the  area  of  a  circle  whose  radius  is  200  ft. 
Solution:  (1)  ^=7rll2^ 

(2)  A  =  3. 1416  X  40000  =  125664. 
.'.  the  area  is  125664  sq.  ft. 

4-.  Find  the  area  of  the  square  that  can  be  inscribed  in  a 
circle  whose  diameter  is  20  ft. 

Note. — A  square  is  inscribed  in  a  circle,  if  all  of  its  angles  lie  upon  the 
circumference.  A  square  is  circumscribed  about  a  circle,  if  it  incloses 
the  circle  with  its  sides  touching  the  circumference. 


ADVANCED    ARITHMETIC. 


281 


Explanation  :  In  the  figure,  OB  and  OC  are 
radii  of  the  circle,  and  are  therefore  each  10 
feet  long.  But  these  radii  are  sides  of  the 
right  triangle  COB,  which  is  one-fourth  of 
the  required  square  ABCD.  Find  the  area 
of  the  triangle  and  multiply  by  4. 

Solution  :   In  the  triangle :  A  =  — - —  =  50. 

:.  the  area  of  square  =  4  x  50  sq.  f t.  =  200 
sq.  ft. 


EXERCISE  Gil. 

1.  Study  and  give  from  memory  the  solutions  of  the  two 
problems  of  this  Article. 

2.  Commit  to  memory  Relations  VII  to  X.  « 
S.    Define  and  draw  a  regular  polygon. 

If.    What  is  a  pentagon?     Octagon?     Decagon? 

5.  Define  a  circle  ;  circumference ;  diameter ;  radius ;  perim- 
eter; apothem. 

6.  What  is  the  formula  for  the  area  of  a  circle  ?  A  regular 
polygon? 

7.  Find  diameter  of  a  circle  whose  area  is  2829.44  sq.  in. 

8.  Find  the  radius  of  a  circle  whose  circumference  is  376.992 
feet. 

9.  Find  the  apothem  of  a  regular  polygon  whose  area  is  20800 
sq.  yd.  and  whose  perimeter  is  520  yd. 

10.  Find  the  area  of  a  circle  whose  circumference  is  87.6992 
ft. 

11.  Upon  how  much  pasture  can  a  horse  graze,  if  he  is  tied 
by  a  rope  60  ft.  long  ? 

12.  A  horse  is  tied  to  the  corner  of  a  house  40  ft.  wide  and  60 
ft.  long,  by  a  rope  90  feet  long.  Over  how  much  ground  can 
the  horse  go  ? 

IS.  Find  one  side  of  the  square  that  can  be  inscribed  in  the 
circle  whose  diameter  is  50  ft. 


282 


ADVANCED    ARITHMETIC, 


^    IJf.    Find  the  area  of  the  circle  if  one  side  of  its  inscribed 
square  is  25  ft. 

15.  Find  the  circumference  of  a  circle  whose  area  is  254.4696 
sq.  ft. 

16.  The  radius  of  a  circle  is  34  inches.     Find  the  area  of  the 
inscribed  square. 

n.   A  circle  is  62.882  ft.  in  circumference.     Find  the  area  of 
the  circumscribed  square. 

3.     SOLIDS. 

127.   Polyhedrons.— Define  solid.     (See  p.  156.) 
A  Polyhedron  is  a  solid  bounded  by  planes.     The  bound- 
ing planes  are  called  faces.     The  lines  of  intersection  of  the 
planes  are  called  edges.     The  points  where  the  lines  of  inter- 
section meet  are  called  vertices. 


A  rectangular  solid. 


A  cube. 


A  Rectangular  Solid  is  a  polyhedron  bounded  by  rec- 
tangular planes. 

A  Cube  is  a  rectangular  solid  whose  dimensions  are  all  equal. 

A  Prism  is  a  polyhedron  two  of  whose  faces 
called  bases  are  parallel  polygons,  and  the  other 
faces,  called  lateral  faces,  intersect  in  parallel  lines. 
The  altitude  of  a  prism  is  the  perpendicular  dis- 
tance between  the  planes  of  its  bases.  If  all  the 
lateral  faces  of  a  prism  are  rectangles,  the  prism  is 
a  right  prism;  if  not,  it  is  an  oblique  prism. 

A  right    triangular 
prism. 

NoTK  1. — We  will  study  only  right  prisms  in  this  book.    The  lateral 
(or  convex)  surface  of  a  prism  is  the  sum  of  the  areas  of  the  lateral  faces  ; 


ADVANCED    ARITHMETIC. 


288 


A  right  pentagonal 
pyramid. 


the  total  surface  is  the  sum  of  the  lateral  area  plus  the  areas  of  the  two 
bases. 

Note  2. — A  prism  with  a  triangle  for  a  base  is  a  triangular  prism  ;  with 
a  square  base,  a  square  prism;  with  a  hexagon  for  a  base,  a  hexagonal 
prism,  etc. 

A  Pyramid  is  a  polygon  one  of  whose 
faces,  the  base,  is  a  polygon,  and  the  lateral 
faces  triangles  meeting  in  a  point  called  the 
vertex  of  the  pyramid.  If  the  lateral  faces  are 
all  equal  isosceles  triangles,  the  pyramid  is  called 
a  right  or  regular  pyramid  ;  if  the  lateral  faces  are 
not  isosceles  triangles,  the  pyramid  is  an  oblique 
pyramid.  The  perpendicular  distance  from  the 
vertex  to  the  base  is  called  the  altitude. 

The  slant  height  of  a  right  pyramid  is  the  distance  from  the 
vertex  to  the  middle  point  of  one  side  of  the  base. 

Note. — A  pyramid  with  a  triangle  for  a  base  is  a  triangular  pyramid  ; 
with  a  square  base,  a  square  pyramid  ;  with  a  pentagon  for  a  base,  a  pen- 
tagonal pyramid,  etc. 

EXERCISE  cm. 

Define  : 

1.  A  solid. 

2.  A  polyhedron. 

3.  A  face  of  a  polyhedron. 

4.  An  edge  of  a  polyhedron. 

5.  A  vertex  of  a  polyhedron. 

6.  A  rectangular  solid, 

7.  A  cube. 

8.  A  prism. 

9.  Bases  of  a  prism. 

10.  The  altitude  of  a  prism. 

11.  A  right  prism. 


12.  An  oblique  prism. 

13.  A  triangular  prism. 
IJ^.  A  square  prism. 

15.  An  octagonal  prism. 

16.  A  pyramid. 

17.  The  altitude  of  a  pyramid. 

18.  A  triangular  pyramid. 

19.  A  pentagonal  pyramid. 

20.  A  right  pyramid. 

21.  Lateral  area. 

22.  Total  area. 


Problem  1:    Letting  1,  w  and   t   rcptresent  length,  loidth  and 


234  ADVANCED    ARITHMETIC. 

thickness  respectively,  develop  the  formula  for  the  volume  of  a  rec- 
tangular solid. 

BY   EQUATION. 

Solution  : 

(1)  Vol.of  asolidZft.  l.,i(;ft.w.,<ft.  th,  =(  )cu.  ft.?    (Question.) 

(2)  Vol.  of  a  solid  1  ft.  1.,  1  ft.  w.,  1  ft.  th.  =  1  cu.  ft.     (Basis.) 
Zx(2)  =  (3)  Vol.  of  a  solid  I  ft.  1.,  1  ft.  w.,  1  ft.  th.  =  /cu.  ft. 

IV  x(3)  =  (4)  Vol.  of  a  solid  Ift.  1.,  w  ft.  w.,  1  ft.  th.  =  lw  cu.  ft. 
<x(4)  =  (5)  Vol.  of  a  solid  I  ft.  1.,  w  ft.  w.,  t  ft.  th.  =  lwt  cu.  ft,  answer. 

BY   PROPORTION. 

Solution  : 

(1)  Vol.  of  a  solid  I  ft.  1.,  w  ft.  w.,  t  ft.  th.  =  (  )  cu.  ft.?     (Question.) 

(2)  Vol.  of  a  solid  1  ft.  1.,  1  ft.  w.,  1  ft.  th.  =  1  cu.  ft.    (Basis.) 

(3)  Ixwyt  :  1x1x1  ::  (  )  !  1? 

...    Ixwxtxl 

(4)  -j^^-j^^-j^    =  Iwt,  answer. 

Using  V  for  volume,  the  result  may  be  written  as  follows ; 

V=lwt. 

Relation  XI.  Abstractly,  the  volume  of  a  rectangular  solid  is 
equal  to  the  continued  product  of  its  length,  width  and  thickness. 

Problem  2:    Develop  the  formula  for  the  volume  of  a  cube. 

Development :  Since  a  cube  is  a  rectangular  solid  whose  length,  width, 
and  thickness  are  all  equal,  each  dimension  in  the  last  formula  may  be 
represented  by  E  (edge).    Then,  the  formula  may  be  written  — 
V=ExE  xE;  or, 

V=EK 
Relation  XII.     Abstractly,  the  volume  of  a  cube  is  equal  to  the 
cube  of  its  edge. 

Problem  3 :  Develop  the  formula  for  the  lateral  or  convex  area 
of  a  right  prism. 

Development :  Since  the  lateral  faces  of  a  right  prism  are  rectangles 
all  having  one  common  altitude,  the  altitude  of  the  prism,  and  the  sum 
of  their  bases  the  perimeter  of  the  base  of  the  prism,  the  sum  of  their 
areas  will  be  the  product  of  these  ;  or, 

L.A.=Pa, 


ADVANCED    ARITHMETIC.  235 

Relation  XIII.  Abstractly,  the  lateral  area  of  a  right  prism 
is  equal  to  the  product  of  the  altitude  and  the  perimeter  of  the  base. 

Problem  4 :  Develop  the  formula  for  the  total  area  of  a  right 
prism. 

Development :  If  the  area  of  each  base  be  represented  by  B,  the  sum  of 
the  areas  of  the  two  bases  added  to  the  lateral  area  will  give  the  total 
area;  or, 

T.  A.=aP-\-2B. 

Relation  XIV.  Abstractly,  the  total  area  of  a  right  prism  is 
equal  to  the  product  of  the  altitude  and  perimeter  of  the  base,  plus 
two  times  the  area  of  the  base. 

Note. — Remember  that  the  area  of  the  bases  of  a  prism  will  have  to 
be  found  by  the  formulas  for  finding  the  areas  of  polygons. 

Problem  5 :  Develop  the  formula  for  the  volume  of  a  right  prism, 
letting  B  represent  the  area  of  the  base  and  a  the  altitude. 

Solution  :  (1)  Vol.  of  a  prism  with  B  sq.  ft.  base,  a  ft.  high  =  (  )  cu.  ft.? 
(Question.) 
(2)  Vol.  of  a  prism  with  1  sq.  ft.  base,  1  ft.  high  =  l  cu.  ft. 
(Basis.) 
J5  =  (2)  =  (3)  Vol.  of  a  prism  with  B  sq.  ft.  base,  1  ft.  high  =  5  cu.  ft. 
ax (3)  =  (4)  Vol.  of  a  prism  with  B  sq.  ft.  base,  a  ft.  high  =  5a  cu.  ft., 
answer. 
Using  V  for  volume,  the  formula  may  be  written— 

V=Ba. 

Relation  XV.     Abstractly,  the  volume  of  a  right  prism  is  equal 
to  the  product  of  the  area  of  the  base  and  the  altitude. 
Note. — Have  the  pupil  solve  the  above  by  proportion. 

Problem  6 :  Develop  the  formula  for  the  lateral  area  of  a  right 
pyramid. 

Development :  In  a  right  pyramid,  the  lateral  area  is  composed  of  equal 
isosceles  triangles  whose  bases  together  form  the  perimeter  (P)  of  the 
base  of  the  pyramid,  and  whose  common  altitude  is  the  slant  height  (s) 
of  the  pyramid.    Therefore, 

T        A  P^ 

L.A.=-. 


286  ADVANCED    ARITHMETIC. 

Relation  XVI.  Abstractly,  the  lateral  area  of  a  right  pyramid 
is  equal  to  the  product  of  the  perimeter  of  the  base  and  the  slant 
height,  divided  by  2. 

Problem  7 :  Develop  the  formula  for  the  total  area  of  a  right 
prism. 

Development :  If  the  area  of  the  base  {B)  be  added  to  the  lateral  area, 
the  sum  will  be  the  total  area,  or, 

T.  A.^'^+B. 

Relation  XVII.  Abstractly,  the  total  area  of  a  right  pyramid 
is  equal  to  the  product  of  the  perimeter  of  the  base  and  the  slant 
height,  divided  by  2,  plus  the  area  of  the  base. 

Relation  XVIII.  Abstractly,  the  volume  of  a  right  pyramid  is 
equal  to  the  product  of  the  area  of  the  base  and  the  altitude,  divided 
by  3. 

Note. — The  development  of  this  relation  is  too  difficult  to  be  given 
before  the  pupil  has  studied  geometry.  It  may  be  expressed  in  the  fol- 
lowing formula : 

Ba 
3 

EXERCISE  OIV. 

1.  Study  until  you  can  solve  from  memory  each  problem  of 
this  Article. 

2.  Commit  to  memory  Relations  XI  to  XVIII. 

3.  Find  the  area  of  a  rectangular  solid  15  ft.  long,  12  ft. 
wide,  and  1^  ft.  thick. 

4..  Find  the  length  of  a  solid  whose  volume  is  1800  cu.  in., 
width  30  in.,  thickness  12  in. 

5.  Find  the  volume  of  a  cube,  one  of  whose  edges  is  46  in. 

6.  Find  one  edge  of  a  cube  whose  volume  is  39304  cu.  in. 

7.  Find  the  volume  of  a  cube  whose  total  area  is  486  sq.  in. 


ADVANCED    ARITHMETIC. 


237 


8.  Find  the  lateral  area  and  total  area  of  a  square  prism, 
each  side  of  the  base  being  7  in.  and  the  altitude  48  in. 

9.  Find  the  volume  of  the  prism  in  No.  8. 

10.  One  side  of  the  base  of  a  square  prism  is  9  in.,  its  volume 
is  1620  cu.  in.     Find  its  total  area. 

11.  A  right  prism  has  for  its  base  a  right  triangle  whose  per- 
pendicular and  hypotenuse  are  respectively  12  in.  and  15  in., 
and  the  altitude  of  the  prism  is  30  in.  Find  the  lateral  area, 
total  area,  and  the  volume. 

12.  A  square  pyramid  has  for  one  side  of  its  base  48  in.  and 
its  altitude  128  in.  Find  the  slant  height,  the  lateral  area, 
total  area,  and  volume. 

13.  A  hexagonal  pyramid  has  for  one  side  of  its  base  12  in., 
the  apothem  of  the  base  10.4  in.  If  the  volume  is  7488  cu.  in., 
find  the  altitude,  slant  height,  total  area. 

1^.  Find  the  distance  from  one  lower  corner  to  the  diagonal 
upper  corner  of  a  cube  whose  edge  is  20  ft.  (See  Exercise  XCIX, 
No.  10.) 

128.  Solid  Having  Curved  Surfaces.  — A 
Curved  Surface  is  a  surface  one  or  both  of  whose  dimen- 
sions change  direction  at  every  point. 

A  Globe  or  Spliere  is  a  solid  bounded 
by  a  curved  surface  all  parts  of  which  are 
equally  distant  from  a  point  within  called 
the  center.  A  straight  line  passing  through 
the  center  of  a  sphere  and  terminating  at 
each  end  in  the  surface  is  a  diameter  of  the 
sphere.  A  straight  line  extending  from  the 
center  to  the  surface  is  a  radius  of  the 
sphere. 

A  cylindrical  surface  is  a  surface  which  uniformly  and  con- 
tinuously changes  the  direction  of  one  and  only  one  of  its 
dimensions. 


A  sphere. 


238 


ADVANCED    ARITHMETIC. 


A  right  cylinder. 


A  Right  Cylinder  is  a  solid  bounded  by 
two  equal  parallel  circles  and  a  cylindrical  sur- 
face joining  the  circumferences  of  these  circles 
at  right  angles  to  the  planes  of  the  circles.  The 
circles  are  called  bases,  and  the  perpendicular 
distance  between  the  bases,  the  altitude. 

If,  at  the  center  of  a  circle,  a  perpendicular 
to   the   circle   be   erected,  and   lines   could   be 
drawn  from' some  point  in  the  perpendicular  to  every  point  in 
the  circumference  of  the  circle,  the  lines  thus   drawn  would 
form  a  continuous  curved  surface.     Such  a  sur- 
face is  called  a   conical   surface.     A   solid   thus 
bounded   by  a  circle  and  a   conical   surface   is 
called  a  Right  Cone.     The  circle  is  called 
the  base;   the  point  at  the  top,  the  apex;   the 
perpendicular   from  the  apex  to  the  base,  the 
altitude;  and  a  straight  line  from  the  apex  to 
the  circumference  of  the  base,  the  slant  height.       a  right  cone. 

Note. — The  author  has  found  it  difficult  to  frame  definitions  for  cylin- 
der and  cone  sufficiently  accurate  to  satisfy  him  and  at  the  same  time 
sufficiently  simple  in  its  wording  to  be  easily  comprehended  by  the  aver- 
age pupil.  The  teacher  should  be  especially  careful  in  explaining  these 
definitions. 

Relation  XIX.  Abstractly,  the  area  of  the  surface  of  a  sphere 
is  equal  to  4  times  3.74-16  times  the  square  of  the  radius  of  the 
sphere. 

Note. — The  development  of  this  relation  is  too  difficult  to  be  given 
before  the  pupil  has  studied  geometry.  It  may  be  expressed  in  the  fol- 
lowing formula : 

Notice  that  this  area  is  just  4  times  the  area  of  a  circle  having  the  same 
radius. 


Problem  1 :     Develop  the  formula  for  the  volume  of  a  sphere. 


ADVANCED    ARITHMETIC.  239 

Development:  A  sphere  is  sometimes  thought  of  as  made  up  of  an  in- 
finite number  of  pyramids,  all  having  their  vertices  at  the  center  of  the 
sphere ;  the  radius  of  the  sphere  for  their  altitudes  ;  and  their  bases  so 
infinitely  small  as  to  form  one  continuous  surface  —  the  surface  of  the 
sphere.  Thus  considered,  the  volume  of  the  sphere  is  the  sum  of  the 
volumes  of  all  these  pyramids,  the  sum  of  whose  bases  is  the  area  {A)  of 
the  surface  of  the  sphere,  and  whose  common  altitude  is  the  radius  {R)  of 
the  sphere.   Then,  ^^_AR 

"    3  * 

But  from  relation  XIX, 

A=AirR^. 


Relation  XX.  Abstractly^  the  volume  of  a  sphere  is  equal  to  If. 
times  3. 14. 16  times  the  cube  of  the  radius,  divided  by  3. 

Problem  2 :  Develop  the  formulas  for  the  right  cylinder. 

Development:  If  we  may  regard  the  circle  as  a  regular  polygon  of  an 
infinite  number  of  sides  (see  problem  2,  p.  229),  then,  the  cylinder  may 
be  considered  as  a  regular  prism  with  an  infinite  number  of  lateral  faces 
so  small  as  to  form  a  curved  surface.  Then,  all  the  formulas  for  the  right 
prism  are  true  for  the  right  cylinder. 

L.  A.=Pa. 
T.  A.=Pa+2  5. 
V=Ba. 

Note. —  In  the  cylinder,  the  base  {B)  is  the  area  of  the  circle  and  the 
perimeter  (P)  is  the  circumference  of  the  circle. 

Relations  :     See  those  for  the  prism  (  Relations  XIII-XV). 

Problem  8.     Develop  the  formulas  for  the  right  cone. 

Development:  The  same  line  of  reasoning  that  makes  the  cylinder  a 
prism  with  an  infinite  number  of  faces,  makes  the  cone  a  regular  pyra- 


240  ADVANCED    ARITHMETIC. 

mid  with  an  infinite  number  of  lateral  faces  so  small  as  to  form  a  curved 
surface.  Then,  all  the  formulas  for  the  right  pyramid  are  true  for  the 
right  cone. 

8  • 

Relations.     See   those   for   the   pyramid   (Relations  XVI- 
XVIII). 

EXERCISE  OV. 

1.  Develop  problems  1,  2  and  3. 

2.  Repeat  the  relations  XIII-XV,  making  necessary  changes 
to  adapt  them  to  the  cylinder. 

3.  Repeat  the  Relations  XVI-XVIII,  making  the  necessary 
changes  to  adapt  them  to  the  cone. 

4.  Commit  to  memory  Relations  XIX-XX. 

5.  Find  the  surface  of  a  globe  whose  radius  is  25  inches. 

6.  Find  the  volume  of  a  globe  whose  diameter  is  15  inches. 

7.  Find  the  radius  of  a  sphere  whose  surface  contains  5026.56 
sq.  ft. 

8.  Find  the  volume  of  the  sphere  in  No.  7. 

9.  Find  the  lateral  area  of  a  cylinder,  if  the  perimeter  of  the 
base  and  the  altitude  are  respectively  18  in.  and  24  in. 

10.  Find  the  total  area  of  a  cylinder  whose  altitude  is  48  in., 
and  the  radius  of  whose  base  is  10  in. 

11.  Find  the  volumes  of  the  cylinders  in  Nos.  9  and  10. 

12.  Find  the  lateral  area  of  a  pyramid,  the  area  of  whose 
base  is  118.0976  sq.  in.  and  whose  volume  is  2261.952  cu.  in. 

13.  Find    the    diameter     of    a    sphere     whose     volume    is 
60104701.6416  cu.  in. 

14..    Find  the  Volume  of  a  cone,  if  the  radius  of  the  base  is 
14  ft.  and  the  altitude  21  ft. 


ADVANCED   ARITHMETIC.  241 

15.  Find  the  whole  area  of  a  cone,  if  the  circumference  of 
the  base  is  100.5312  in.  and  the  slant  height  is  86  in. 

16.  Find  the  altitude  in  No.  15. 

17.  Find  the  lateral  area  of  a  c6ne  whose  slant  height  is  21 
in.  and  whose  base  contains  254.4096  sq.  in. 

18.  Find  the  volume  of  the  cone  in  No.  17. 

19.  Find  one  side  of  the  cube  that  can  be  inscribed    in  a 
sphere  whose  diameter  is  20  ft. 

Note. — Compare  this  problem  with  your  solution  of  No.  14,  Exercise 
cm.    The  process  here  will  be  found  to  be  the  reverse  of  that. 

129.     Similar  Figures. — Geometrical   figures   which 
have  the  same  form  are  called   Similar   Figures.      All 

squares  are  similar,  all  equilateral  triangles  are  similar,  all  circles 
are  similar,  all  globes  are  similar. 

Note. —  When  the  figures  to  be  studied  in  a  problem  are  similar,  the 
problem  will  usually  so  indicate. 

Principles  IN  Similar  Figures  :  1.  Corresponding  linear  parts 
are  proportional. 

2.  Areas  are  proportional  to  the  squares  of  corresponding  linear 
parts. 

3.  Volumes  are  proportional  to  the  cubes  of  corresponding  linear 
parts. 

To  illustrate :  If  in  two  circles  their  radii  be  represented  by  R  and  r, 
their  circumferences  by  C  and  c,  and  their  areas  by  A  and  a,  then, 
from  prin.  1,  C  :  c  ::  R  :  r,  and 
from  prin.  2,  A  :  a  ::  C^  :  c^;  or, 
A  :a::  R^  :rK 

If   in  two  spheres  their  diameters  be  represented  by  I)  and  d,  their 
surfaces  by  .1  and  a, and  their  volumes  by  T''and  v,  then, 
from  prin.  2,  A  :  a  ::  I)^  :  d^,  and 
from  prin.  3,   V  :  v  ::  D^  :  d^. 


242  ADVANCED   ARITHMETIC. 

EXAMPIiES. 

1.  If  a  yardstick  standing  upright  casts  a  shadow  of  2|  ft., 
how  high  is  a  flagpole  whose  shadow,  at  the  same  time,  is  75  ft.? 

Note. — The  yardstick  (3  ft.)  and  the  flagpole  are  the  perpendiculars, 
and  the  two  shadows  the  bases  of  two  similar  right  triangles. 

Solution:     (1)  3:  P: :  2^:  75. 
(2)   i^  =  ^  =  90. 
.-.   the  flagpole  is  90  ft.  high. 

2.  A,  who  is  6  ft.  tall,  has  a  statue  80  ft.  tall.  (1)  If  in  the 
statue  the  foot  is  5  ft.  long,  how  long  is  A's  foot?  (2)  A's 
forefinger  is  4  in.  long;  how  long  is  the  corresponding  finger  in 
the  statue? 

Note. — The  statue  and  the  man  are  similar  figures. 

(1)  Solution:  (1)  6:30: :  F:  5. 

(2)  fJ-^  =  \. 
30 

.•.  A's  foot  is  1  ft.  long. 

(2)  Solution:  (1)  6:  30::4: /. 

(2)  /  =  ^  =  20. 

.*.  the  finger  of  the  statue  is  20  in.,  or 
1  ft.  8  in.  long. 

3.  There  are  two  statues  of  the  same  shape;  the  height  of 
one  is  4  ft.  and  of  the  other  16  ft.  If  it  cost  50/  to  paint  the 
first,  what  will  it  cost  to  paint  the  other? 

Note. — Similar  surfaces  are  proportional  to  the  squares  of  their  cor- 
responding linear  parts. 


Solution:   (1)  42:162::  .5:C. 

(2)  C=——  =  S. 
16 

.'.  the  required  cost  is 


ADVANCED   ARITHMETIC.  243 

If..  The  statues  in  Example  3  are  made  of  the  same  kind  of 
material:  if  the  material  for  the  larger  one  cost  $256,  what 
did  the  material  for  the  smaller  cost  ? 

Note. —  The  prices  are  proportional  to  their  volumes,  and  from  prin- 
ciple 3,  to  the  cubes  of  their  corresponding  linear  parts. 

Solution:  (1)  4^:  IG^: :  C:  256. 
64X256^^ 
^  '  3896 

.".  the  required  cost  is  $4. 

5.  A  20-inch  water-pipe  branches  into  3  equal  pipes :  what 
must  be  the  diameter  of  each,  that  they  together  may  convey 
the  same  water  as  the  original  pipe  ? 

Note.— The  area  of  the  opening  in  each  small  pipe  must  be  i^  of  that 
of  the  large  pipe.  Areas  of  the  circular  openings  are  proportional  to  the 
squares  of  their  diameters. 

Solution:  (1)  2(^:  D^::l:h 

(2)  D^  =  HK 

(3)  D  =  11.547-. 

.*.  the  area  of  each  small  pipe  must  be  11.547 -in. 

6.  How  many  balls  2  inches  in  diameter  will  equal  in  vol- 
ume one  2  feet  in  diameter  ? 

Note. — Balls  are  similar  figures. 

Solution:  (1)  2  ft.  =  24  in. 

(2)  F:v:: 243:23. 

(3)  v='^^  =  msv. 

,*.  it  will  require  1728  small  balls. 
EXERCISE  CVI. 

1.  The  diameters  of  two  circles  are  to  each  other  as  3:4; 
the  circumference  of  the  first  is  42  ft. :  find  the  circumference 
of  the  second. 

2.  How  many  pipes  2  in.  in  diameter  will  carry  as  much 
water  as  3  pipes  6  in.  in  diameter,  neglecting  the  friction  ? 


244  ADVANCED    ARITHMETIC. 

3.  Two  men  buy  a  grindstone  2  ft.  in  diameter;  the  first 
grinds  till  the  stone  is  18  in.  in  diameter.  How  much  should 
each  pay,  allowing  a  circular  hole  in  the  center  8  in.  in  diam- 
eter, which,  if  it  were  counted  as  stone,  would  be  worth  2/  ? 

J/..  A  and  B  buy  a  ball  of  twine  6  in.  in  diameter  for  16/.  A 
winds  off  the  twine  till  the  ball  is  but  3  in.  in  diameter,  and 
then  gives  the  ball  to  B.     How  much  should  each  pay  ? 

5.  How  many  leaden  balls  ^  in.  in  diameter  can  be  made 
from  a  leaden  ball  2  in.  in  diameter  ? 

6.  The  diameter  of  Jupiter  is  11  times  that  of  the  earth:  (1) 
Its  surface  is  how  many  times  that  of  the  earth  ?  (2)  Its  vol- 
ume is  how  many  times  that  of  the  earth  ? 

130.  Review^. — 

EXERCISE  CVII. 

1.  Find  the  hypotenuse  of  a  right  triangle,  if  its  base  and 
perpendicular  are  36  ft.  and  48  ft.,  respectively. 

2.  Find  the  base  of  a  right  triangle,  if  the  hypotenuse  and 
perpendicular  are  100  ft.  and  80  ft.,  respectively. 

S.  The  base  and  perpendicular  of  a  right  triangle  are  equal, 
and  the  hypotenuse  is  125  ft. :  find  the  perpendicular. 

Jf..  Find  the  length  of  one  side  of  the  square  that  can  be  cut 
from  a  circle  whose  diameter  is  250  inches. 

5.  How  long  is  the  ladder  that  can  stand  27  ft.  from  a  wall 
86  ft.  high,  and  reach  the  top  of  it  ? 

6.  Find  the  area  of  a  triangle  whose  base  and  altitude  are 
180  ft.  and  75  ft.,  respectively. 

7.  Find  the  area  of  a  right  triangle  whose  base  and  perpen- 
dicular are  80  ft.  and  40  ft. ,  respectively. 

Note. —  In  a  right  triangle  the  base  and  perpendicular  may  be  con- 
sidered the  base  and  altitude. 

8.  The  area  of  a  right  triangle,  whose  base  and  perpendicular 
are  equal,  is  1800  sq.  ft. :  find  the  length  of  the  base. 


ADVANCED    ARITHMETIC.  245 

9.  The  area  of  a  triangle  is  1200  sq.  ft.,  its  altitude  is  30  ft. : 
find  the  length  of  its  base. 

10.  Find  the  area  of  the  inner  surface  of  the  walls  of  a 
house,  if  the  dimensions  are  80  ft.  by  20  ft.  by  12  ft.  in  the 
clear. 

11.  Subtract  from  the  result  found  in  No.  10  the  area  of  6 
windows  6  ft.  by  2  ft.  3  in.,  and  3  doors  6  ft.  8  in.  by  3  ft.  4  in. 

12.  A  field  is  in  the  shape  of  a  parallelogram  whose  base  is 
50  rd.  and  whose  altitude  is  20  rd. :  find  how  many  acres  it 
contains. 

13.  A  rectangular  field  has  one  side  twice  as  long  as  the 
other;    its  area  is  20  acres:    find  the  length  of  its  sides. 

Note.— Reduce  the  acres  to  square  rods. 

14^.  The  length  of  a  rectangular  field  is  to  its  width  as  8  to  5, 
and  it  contains  25  acres :  find  its  sides. 

15.  A  trapezoid  has  for  its  bases  49  ft.  and  71  ft.,  and  its 
altitude  is  50  ft. :    find  its  area. 

16.  A  pasture,  in  the  form  of  a  trapezoid,  contains  1200 
acres,  its  parallel  sides  are  2  mi.  and  3  mi.  long:  how  far 
across  it? 

17.  In  a  trapezoid  the  bases  are  to  each  other  as  2  to  3,  the 
altitude  is  14  ft.,  and  the  area  is  350  sq.  ft. :  find  the  length  of 
the  bases. 

18.  The  circumference  of  a  circle  is  326.72536  ft. :  find  the 
radius,  diameter,  and  area. 

19.  The  area  of  a  circle  is  45238.896  sq.  ft. :  find  the  radius 
and  circumference. 

20.  How  long  must  a  rope  be  to  fasten  a  horse  so  that  he 
may  graze  upon  -J  acre  and  no  more  ? 

21.  Two  sides  of  a  rectangle  are  30  ft.  and  40  ft. :  find  the 
area  of  the  circumscribed  circle. 

Note. —  The  diameter  of  the  circle  is  the  diagonal  of  the  rectangle. 


246  ADVANCED   ARITHMETIC. 

22.  A  circle  is  circumscribed  about  a  rectangle  whose  sides 
are  to  each  other  as  3  to  4 :  find  the  area  of  the  rectangle  if 
the  diameter  of  the  circle  is  60  ft. 

23.  Find  the  number  of  cubic  inches  in  a  bushel  measure,  a 
right  cylinder  whose  altitude  is  8  inches  and  the  radius  of 
whose  base  is  9^  inches. 

2^.  What  will  it  cost  to  paint  the  convex  surface  of  a  cylin- 
drical tank  8  ft.  high  and  14  ft.  across  the  bottom,  if  the 
painter  charges  15|/  i^er  sq.  yd.? 

25.  How  many  gallons  of  water  will  the  tank  described  in 
No.  24  hold  ?     (Consider  the  dimensions  inside  measure.) 

26.  Find  the  volume  of  a  rectangular  solid,  if  its  base  is  a 
rectangle  10  in.  by  25  in.  and  its  altitude  20  in. 

27.  Find  the  surface  of  the  solid  in  No.  26. 

28.  Find  the  surface  and  the  volume  of  a  cube,  if  its  altitude 
is  17  ft. 

29.  The  apothem  of  the  base  of  a  right  hexagonal  pyramid  is 
8  in.,  the  length  of  one  side  of  the  base  is  9.2876  in.,  and  the 
altitude  of  the  pyramid  is  8  ft.     Find  the  volume. 

30.  Find  the  whole  surface  of  the  pyramid  in  No.  29. 

31.  Find  the  volume  of  a  right  cone  the  diameter  of  whose 
base  is  18  ft.  and  whose  altitude  is  80  ft. 

32.  Find  the  surface  of  the  cone  in  No.  81. 

33.  The  circumference  of  a  sphere  is  188.4954  in. :  find  its 
vqlume  and  surface. 

3J^.  The  diameter  of  a  sphere  is  48  in. :  find  its  surface  and 
volume. 

35.  The  volume  of  a  sphere  is  1767.15  cu.  in. :  find  its  radius, 
diameter,  circumference,  and  surface. 

36.  The  surface  of  the  earth  is  196668855.7504  sq.  mi. :  find 
its  volume. 

37.  The  bases  of  two  similar  triangles  are  respectively  10  in. 
and  16  in.  If  the  altitude  of  the  smaller  is  9  ft.  2  in.,  find  the 
altitude  of  the  larger. 


ADVANCED    ARITHMETIC.  247 

B.     FEBCENTAGE. 

1.  THE  PERCENT  AGE  FORMULA. 

131.  Elements. —  There  are  three  elements  in  percent- 
age problems : 

(1)  A  number  some  per  cent  of  which  is  to  be  considered.  This 
number  is  called  the  Base;  and,  in  the  percentage  formula  to 
follow,  it  is  represented  by  B. 

(2)  A  number  of  per  cent  of  the  base.  This  number*  is  repre- 
sented in  the  formula  by  p. 

(8)  The  number  which  equals  the  j)er  cent  of  the  base  considered. 
This  number  is  called  the  percentage ;  and,  in  the  formula,  is 
represented  by  P. 

To  illustrate : 

5%of240isl2. 

5  is  the  number  of  per  cent,  p. 

240,  is  the  base,  B. 

12  is  the  percentage,  P. 

Many  authors  speak  of  two  additional  elements.  Amount  and 
Difference.     For  example : 

25%  more  than  cost  =  100%  of  cost+25%  of  cost.     (Amount.) 
25%  less  than  cost  =  100%  of  cost  — 25%  of  cost.    (Difiference.) 

It  is  not  necessary  to  put  any  special  study  or  emphasis  upon 
these  as  elements. 

25%  more  than  a  number =125%  of  that  number. 
25%  less  than  a  number  =  75%  of  that  number. 


♦This  number  has  received  the  name  Bate,  and  in  the  formulas  of  most  arithmetics  is 
represented  by  R.  It  is  probable  that  it  has  been  so  called  from  the  fact  that  this  number  is 
often  a  rate.  For  example :  Rates  of  interest,  rates  of  commission,  rates  of  gain  or  loss,  rates 
of  insurance,  are  usually  expressed  in  per  cent.  But  when  applied  to  the  simple  percentage 
problem,  the  word  rate  is  a  misnomer.  One  number  may  be  some  per  cent  of  another,  but  not 
some  rate  of  another. 


248  ADVANCED   ARITHMETIC. 

EXERCISE  OVIII. 

Recite  this  exercise  orally  : 

i.  50%  more  than  cost=(   )%  of  cost  ? 

2.  94%  more  than  my  age  =  (   )%  of  my  age  ? 

3.  47%  less  than  $500  =(   )%  of  $500? 

4..  35%  less  than  the  price=(   )%  of  the  price  ? 

5.  148%  of  $900  =(   )%  more  than  $900  ? 

6.  66|%  of  $600=  (   )%  less  than  $600? 

7.  100%  more  than  a  number  =(   )%  of  that  number  ? 

8.  100%  less  than  a  number  =(   )%  of  that  number  ? 

9.  500%  of  a  number=(   )%  more  than. that  number? 

132.    Developing  tlie  Percentage  Formiila. — 

The  percentage  formula  is  the  equation  which  relates  the  three 
elements,  base,  per  cent,  and  percentage.  It  may  be  developed 
either  by  the  equation  method  or  the  proportion  method  of  so- 
lution. 

Problem:  Find p%  of  B. 

Note. — We  know  that  100^  of  anything  is  all  of  it;  then,  we  know  that 
100%  of  B  =  B. 

BY    EQUATIONS. 

Solution :  (1)  p%otB  =  {  )  ?    (Question.) 
(2)  100^  of  5  =  5.    (Basis.) 

T^of(2)  =  (3)  l%ot  B  =  ^ 


px(3)  = 

=  (4)p^  of  5=^,  answer. 

Solution : 

BY    PROPORTION. 

(1)  p%otB  =  {  )?    (Question.) 

(2)  100^  of  5  =  5.    (Basis.) 
(3)p:100::(  ):5. 

But,  since  this  answer  represents  the  percentage,  P,  we  have — 

lOO* 


ADVANCED   ARITHMETIC.  249 

Relation  :     Abstractly,  the  percentage  is  equal  to  the  product 
of  the  base  and  the  number  of  per  cent,  divided  by  100. 

EXAMPIjBS. 

1.  Find  8%  of  $50. 

Solution :  (1)  P=  ^. 

^^^  ^      100     *■ 
Answer,  $4. 

2.  $20.80  is  4%  of  what  number  ? 

Solution:  {1)^  =  20.8.    (Why?) 

(2)  ^  =  ^^^^  =  520. 
4 

Answer,  $520. 

3.  What  %  of  $600  is  $540  ? 

,SoZM<tori:  (1)  ^=|40.     (Why?) 

540xl00_ 
^^^^'-      600      "^• 
Answer,  90^. 

4-.  600  is  50%  more  than  what  number  ? 

Note. — bO%  more  than  a  number  is  150^  of  that  number. 

Solution:  (1)  -^^  =  600. 

._,   ,,    600x100     .^ 

(2)  ^  =  — iKA     =400,  answer. 

5.  520  is  how  many  %  less  than  650  ? 
Solution:  (1)  ^^  =  520. 

Answer,  20^  less.    (Why  ?) 


250  ADVANCED   ARITHMETIC. 

6.  $600  is  20%  less  than  what  number  ? 
Note.— 20^  less  than  a  number  is  80^  of  it. 

Solution:  (1)  ^  =  600. 

,2)  B  =  822AWO^,5o 

oU 

Answer,  $750. 

7.  Find  a  number  B,  if  p%   of  it  is  P.     (By  the  equation 
method.) 

Solution:  (1)  100^  of  5  =  (  )  ?     (Question.) 
{2)p%oiB  =  P.    (Basis.) 

(2)-Hp  =  (3)  \%otB=-. 

100  P 
100x(3)  =  (4)  100^  of  B  = ;or, 

5  = ,  result. 

P 

Note.— Let  the  pupil  give  the  relation. 

EXERCISE  CIX. 

1.  Commit  to  memory  the  relation  given  on  p.  249. 

2.  Develop  from  memory  the  percentage  formula. 

3.  Show  how  formula  in  example  7  can  be  obtained  directly 
from  the  percentage  formula. 

4.  P  is  how  many  %  oi  B? 

100  P 
Note. — The  solution  of  number  4  gives  the  formula  jj  =  — =— 

5.  Show  that  p = — ^—  can  be  obtained  directly  from  the  per- 
centage formula. 

6.  $54  is  9%  of  what  ? 

7.  Find  47%  of  $460. 

8.  56  is  what  %  of  64  ? 


ADVANCED   ARITHMETIC.  251 

9.   564  is  20%  more  than  what  ? 

10.  376  is  20%  less  than  what  ? 

11.  What  number  added  to  20%  of  itself  gives  564  ? 

Note. — Show  that  Nos.  9  and  11  are  the  same. 

12.  What  number  less  20%  of  itself  gives  876  ? 
Note.—  Show  that  Nos.  10  and  12  are  the  same. 

13.  What  number. added  to  16|%  of  itself  gives  770? 

14.  What  number  less  58%  of  itself  gives  105? 

15.  $860  is  how  many  %  of  $600  ? 

16.  f  is  how  many  %  more  than  \  ? 

17.  \  is  how  many  %  less  than  |  ? 

18.  5%  of  420  is  how  many  %  less  than  85  ? 

19.  $35  is  how  many  %  more  than  7%  of  $400  ? 

20.  4!d(fo  of  $500  is  how  many  %  of  f  of  $800  ? 

2.     PEBCENTAQE  WITHOUT  TIME. 

133.  Profit  and  Loss. 

Principles:  1.  The  gain  or  loss  is  some  number  of  per  cent  of 
the  cost  price. 

2.  The  selling  price  is  equal  to  the  cost  price  plus  the  gain  or 
minus  the  loss. 

TERMS. 

C,  cost  price.  L,  Loss. 

S,  selling  price.  p,  number  of    %  of   gain  or 

(t,  gain  or  profit.  loss. 

Development  of  formulas  for  Profit  and  Loss  : 

1.  ^=^'  (Prin.  1.) 

2.  2>  =  ^.  (Prin.l.) 


252  ADVANCED   AKITHMETIC. 

Relation  I.  Abstractly,  the  gain  or  loss  is  equal  to  the  product 
of  cost  price  and  the  number  of  per  cent  of  gain  or  loss,  divided  by 
100. 

When  there  is  a  gain, 

(1)  S=C+G.  (Prin.  2.) 

But,  (2)  G  =  ^ 
Then,(3)^'=C+^  =  150^;or, 

0(100+p) 

^'  ^- — 100^  • 

When  there  is  a  loss, 

(1)  S=C-L.  (Prin.  2.) 

But,  (2)  L=P^. 
100 

O(lOO-y) 

^-  *- — ioo~- 

Relation  II.  Abstractly,  the  selling  price  is  equal  to  the  product 
of  the  cost  price  and  100  plus  the  per  cent  of  gain,  divided  by  100 ; 
or  the  product  of  the  cost  price  and  100  minus  the  per  cent  of  loss, 
divided  by  100. 

EXAHPIiES. 

1.  How  much  does  a  man  gain  by  selling  an  article  that  cost 
$12.50  at  40%  profit  ? 

Solution:  ^,,40x12.5^^      (Why?) 
»        100 

Answer,  $5. 

2.  An  article  cost  $4.20 :  for  how  much  must  it  be  sold  to 
gain  20%  ? 

Solution:   ^  =  4.2x120^^  q^      (Why?) 
100 

Answer,  $5.04. 


ADVANCED    ARITHMETIC.  258 

3.  A  drover  paid  $1820  for  cattle  that  he  afterwards  sold  at  a 
loss  of  16|%.     What  did  he  lose  ? 

Solution:   L  =  M^il??2  =  220.     (Why?) 
100 

Answer,  $220. 

Note. — Use  cancellation  when  practicable.     16|  is  contained  in  100  6 
times. 

4„  Flour  that  cost  $4.50  per  barrel  sold  for  $4.95  per  barrel. 
What  was  the  per  cent  of  gain  ? 

•  Solution:   (1)  ^  =  ^iM±£) 

100 

(2)  4.95  =  i:511?2±£> 

(2)  =  (3)M1100+£)  =  4.95. 
'  ^    '  '  100 

100  X  (3)  =  (4)  450+4.5j9  =  495. 

Transpose,  (5)  4.5  p  =  495 -450  =  45. 
(5) -H  4.5  =  (6)  p  =  10. 

Answer,  W. 

Another  Solution  :  (!)  G  =  S-C  =  $4.95 - $4.50  =  $.45. 

,3)M^  =  .45. 

100  X  (3)  =  (4)  4.5p  =  45. 
(4) -=-4.5  =  (5)  p  =  10. 

Answer,  W. 

5.  Find  the  per  cent  of  loss,  if  goods  that  cost  $2720  sell 
for  $2380. 

Solution:   {1)S  =  ^^^^^^ 
100 

(2)  2380  =  ?™aOO-p) 

'  100 

100  X  (2)  =  (3)  238000  =  272000 -2720  p. 

Transpose ,  (4)  2720  p  =  272000  -  238000  =  34000. 

(4) -H  2720  =  (5)  «  =  ?155^=12k 
^      2720         ^ 

Answer,  121^. 


254  ADVANCED    ARITHMETIC. 

Another  Solution :  (1)  L  =  C-S  =  $2720  -  $2380  =  $340. 

(2)  ^  =  L. 
100 

(3)  2720^^340, 

100 

(4)  2720  p  =  34000. 

(5)  «  =  ?^??9  =  12i. 
^  '  ^     2720 

Answer,  12^5^. 

Note.— In  No.  4,  first  solution,  equation  (2)  turned  around  gives  equa- 
tion (3).  If  I  had  transposed,  I  would  have  a  minus  sign  before  the  p. 
In  No.  5,  first  solution,  I  transposed  in  equation  (3)  to  get  equation  (4). 
If  I  had  turned  the  equation  around  as  in  No.  4,  I  would  have  a  minus 
sign  before  the  p.  If  the  letter  has  a  plus  sign  in  the  right  member,  turn  the 
equation  around,  to  change  to  tJie  left  rnemher;  if  the  letter  has  a 
minus  sign,  transpose,  to  get  rid  of  the  minus  sign. 

6.  A  merchant  bought  goods  at  20%  less  than  market  value 
and  sold  them  at  20%  more  than  market  value.  What  %  did 
he  gain? 

Solution :  (1)  C  =  80^  of  market  value.     (Why  ?) 
(2)  G^  =  40^  of  market  value.    (Why  ?) 

Note. — The  problem  then  becomes,  "40^  is  what  %  of  80^  ?  "    ' 

,-,  40x100    „ 

Answer,  50^. 

Oral  Solution :  Since  the  cost  is  80^  of  market  value,  and  the  gain  is 
40^  of  the  market  value,  the  gain  is  one-half,  or  50%  of  the  cost. 

7.  A  man  lost  16|%  by  selling  a  house  for  $538  less  than  it 
cost.     Find  the  cost. 


ADVANCED   ARITHMETIC.  255 


Solution:  (1)  L  =  ^--. 

(2)538=^. 

(3)  ^^538x100^3^ 
16| 


Answer, 

8.  A  merchant  sold  goods  for  $980.28,  thereby  losing  10% . 
Find  the  cost. 

Solution:  {I)  S=^^^-^. 

(2)980.28-        ^^        -^^. 

(3)  0  =  ^-55:^^=1089.2. 
Answer,  $1089.20. 

9.  How  must  I  mark  a  pair  of  shoes  that  cost  $2.40,  to  gain 
25%  ?     (  Key  :   Come  and  buy.) 

_,  .    .        _,    2.4  (100+25)    _ 

Solution:  S  = ^- ^  =  3. 

100 

Answer,  m.yy. 

Note. — Keys  are  employed  by  many  merchants  to  mark  the  selling 
price  as  well  as  the  cost  price  of  their  goods,  in  a  manner  known  to 
themselves  only.  The  letters  of  the  key,  taken  in  order,  represent  the 
figures : 

come      and      buy 
1234      567       890 
Then,  ^3.00  =  m.yy.    A  separate  character,  as  x,  is  often  used  instead  of 
repeating  a  letter.    Thus  m.yy  may  be  written  m.yx. 

EXERCISE  ex. 

1.  What  is  the  base  in  profit-and-loss  problems? 

2.  Commit  to  memory  relations  I  and  II. 

3.  Give  from  memory  the  formulas  under  relation  I. 
4-.  Develop  all  the  formulas  for  profit  and  loss. 


256 


ADVANCED    ARITHMETIC. 


No. 
5. 

$C 

P% 

$G 

#L 

$S 

484 

25 

? 

? 

6. 

564 

25 

? 

? 

7. 

840 

33i 

? 

? 

8. 

420 

m 

? 

? 

9. 

248 

80 

m 

? 

? 

10. 

590 

? 

59 

? 

11. 

780 

? 

219 

? 

12. 

48 

50 

? 

? 

58 

20 

13. 

32 

40 

? 

? 

25 

92 

u. 

? 

? 

8 

58 

65 

78 

15. 

? 

? 

2 

50 

10 

16. 

? 

18i 

6 

30 

? 

17. 

? 

6i 

27 

30 

? 

18. 

1  ? 

120 

192 

? 

19.  Jones  sells  goods  costing  him  $1500  at  a  profit  of  16f  %. 
Find  the  gain. 

20.  I  sold  100  hogs,  averaging  300  lb.  each,  at  a  profit  of 
33^%.     If  they  cost  me  1^1200,  find  the  selling  price  per  lb. 

21.  What  was  the  loss,  if  a  pair  of  shoes  that  cost  $2.50  sold 
at  20%  loss? 

22.  How  would  you  mark  an  article  that  cost  $3.20  to  sell 
at  a  20%  profit  ?     (  Key :  Buy  for  cash.) 

23.  If  $187.50  is  gained  on  goods  that  cost  $1250,  find  the 
selling  price,  and  per  cent  of  gain. 

24..  If  $260  is  lost  on  goods  that  cost  $1300,  find  the  selling 
price  and  per  cent  of  loss. 

25.  If  goods  sold  at  16f  %  profit  for  $1085,  find  the  cost 
price. 

26.  If  goods  sold  at  25%  loss  for  $480,  find  the  cost  price. 

27.  Sold  a  horse  for  $80,  losing  20%  ;  with  the  $80  bought 
another,  and  sold  him,  gaining  20%.  Find  the  loss  on  the 
two  transactions. 

28.  A,  failing  in  business,  paid  B  $1512,  which  is  only  60/ 
on  the  $1  (60%).     How  much  did  A  owe  B  ? 


ADVANCED    ARITHMETIC.  257 

29.  Mr.  Smith  bought  goods  and  sold  them  at  a  loss  of  12^%, 
losing  $187.     Find  the  selling  price. 

30.  I  bought  a  horse,  and  afterwards  sold  him  at  a  loss  of 
20%  ;  I  added  20%  to  what  I  then  had,  and  with  this  amount 
bought  another  horse.  I  sold  the  second  horse  at  a  profit  of 
20%  for  $144.     What  did  the  first  horse  cost  ? 

134.  Trade  Discount. — Wholesale  merchants  usually 
mark  the  selling  price  of  their  goods  so  that  they  can  make  a 
reduction  from  this  price  in  selling  to  their  customers.  The 
marked  price  is  often  spoken  of  as  list  price.  The  reduction 
from  the  list  price  is  called  Trade  Discount. 

Note. — Discounts  are  made  for  one  or  more  of  the  following  reasons: 
(1)  The  goods  may  be  marked  in  the  first  place  with  the  expectation  of 
making  a  uniform  reduction  to  all  retail  customers,  (2)  for  large  purchas- 
ers, and  (3)  for  cash. 


"  25%  off,"  or  "25  off,"  means  25%  less  than  the  list  price. 
**  25  and  10  off  "  means  25%  less  than  list  price,  and  then  10% 
less  than  the  amount  left  after  the  first  reduction.  "  J  and  5 
off,"  means  the  same  as  "  88^  and  5  off." 

Principles  :  1.  The  first  discount  is  some  number  of  per  cent 
of  the  list  price. 

2.  Each  succeeding  reduction^  or  discount^  is  some  number  of  per 
cent  of  the  amount  left  after  making  the  last  preceding  reduction. 

3.  The  selling  price ^  after  any  discount^  is  equal  to  the  price  be- 
fore the  discount  J  minus  the  discount. 

TERMS. 

L,  list  price. 

Z),  discount. 

S,  S\  S^\  etc.,  successive  selling  prices. 

Pj  number  of  %  off. 


258  ADVANCED    ARITHMETIC. 

Development  of  the  formulas  for  Trade  Discount. 

1.  D=  ^.  (Prin.l.) 

100 

Relation  I.     Abstractly,   the  discount  is  equal  to  the  product 

of  the  list  price  and  the  number  of  per  cent  of  discount,  divided 

by  100. 

(1)   S  =  L-D.    (Prin.  3.) 

^     jPL    lOOL-pL 

2.  ^.^(1^-^) 


100 

Relation  II.  When  there  is  one  reduction,  abstractly,  the  sell- 
ing price  is  equal  to  the  product  of  the  list  price  and  100  minus 
the  number  of  per  cent  of  reduction,  divided  by  100. 

When  there  are  several  reductions,  the  amount  left,  or  S,  of  the  first 
reduction  becomes  the  L  for  the  next  reduction,  and  so  on. 

Then,  if  another  per  cent  {p^)  be  taken  off  the  new  selling  price, 

S.=  «ii^).    (Prin.  2.) 

Putting  in  place  of  S  its  value  from  the  formula  above,  we  have: 

X(100-p)(100^p-) 
100X100 

In  the  same  way,  a  formula  for  any  number  of  reductions  may  be 
obtained. 

Relation  III.  After  any  number  of  reductions,  abstractly, 
the  selling  price  is  equal  to  the  fraction  whose  numerator  is  the 
product  of  the  list  price  and  the  several  remainders  found  by  sub- 
tracting each  number  of  per  cent  from  100,  and  whose  denominator 
is  the  product  of  as  many  lOO^s  as  there  are  reductions. 


ADVANCED   ARITHMETIC.  259 

EXAMPIiES. 

1.  Find  a  discount  of  20%  off  of  a  list  price  of  $540. 

Solution:  D  =  ^^^  =  m. 
Answer,  $108. 

2.  If  the  list  price  is  $460,  find  the  selling  price  at  30  off. 

Solution:  S  =  ^^^  =  322.     (Why?) 
Answer,  $322. 

3.  If  the  list  price  is  $500,  find  the  selling  price  at  20  and  10 

off. 

^  -  ^.        ^,    500x80x90    -^ 
Solution:  S^  =    ^^^^^    =360. 

Answer,  $360. 

4.  I  sold  an  article  that  cost  me  $12,  at  i,  15,  and  5  off. 
Find  the  selling  price. 

^  ,  ^.         ^,    12x661x85x95    ^  ,„ 
Solution:  S^  =   .^    '        ..^  =6.46. 
100  X 100  X 100 

Answer,  $6.46. 
NoTB.— 33i  will  cancel  66|  and  100. 

5.  If  an  article  sells  at  80  off  for  $14,  find  the  list  price. 

Solution:  {1)^^^^^^^  =  S. 

Answer,  $20. 

6.  If  an  article  sells  at  30  and  10  off  for  $6.30,  find  the  list 

price. 

c  7  ,•       ,,.  xaoo+£Mioo+|/)    ^ 
^^^"'*"^'"  ^^^ looTioo ='^- 

70X90XX 
^^^    100x100  -^•^^- 
,„.    ^     6.30x100x100    ,_. 
^^^^-        70^^90         -^^• 

Answer,  $10. 


^60 


ADVANCED    ARITHMETIC. 


7.  An  article,  listed  at  $20,  sold  at  (   )  and  10  off  for  $13.50. 
Fill  the  blank.  » 

c  7  4-        /l^  lOKA    20(100-p)x90 
^otoon:  (I)  13.50=       ^^/^^      . 

10000x(l)  =  (2)  135000  =  1800(  100 -p). 

(3)  135000  =  180000 -1800  p. 
Transpose,  (4)  1800^-180000-135000. 

(5)  1800^  =  45000. 

.^.         45000    ^^ 

^'^^=1800=^- 
Answer,  2b%. 

EXERCISE  CXI. 

1.  What  does  "  12i  and  10  off  "  mean  ? 

2.  Give  the  principles  governing  Trade  Discount. 

3.  Develop  the  formulas  and  commit  to  memory  the  rela- 
tions. 


No. 

$L 

'P% 

$D 

$  S 

^. 

580 

10,  10 

? 

? 

5. 

460 

? 

69 

? 

6. 

500 

20,  ? 

? 

375 

7. 

? 

15,  10 

9 

459 

8. 

720 

?  20 

? 

432 

9. 

? 

15 

63 

81 

? 

10. 

? 

1 

? 

304 

04 

1216 

16 

ii.  A  man  sold  goods  listed  at  $437  at  20%  off.      What  did 
they  sell  for  ? 

12.  Goods  sold  at  25  and  5  off  of  a  list  price  of  $540.     Find 
the  amount  of  the  discount. 

13.  Goods  marked  at  $380  sold  at  i,  10,  and  10  off.    Find  the 
selling  price. 

11^.  Merchandise  sold  at  20  and  15  off  for  $1360.     Find  the 
list  price. 


ADVANCED   ARITHMETIC.  261 

15.  I  bought  an  article,  marked  it  at  $20,  and  sold  it  at  80 
and  (   )  off  for  $12.60.     Fill  the  blank. 

16.  How  must  I  mark  goods  that  cost  me  $450  so  that  I  may 
sell  10  off  of  list  price  at  a  gain  of  20%. 

17.  I  sold  an  article  at  20  and  10  off.  If  the  discounts 
amount  to  $3.50,  for  how  much  did  the  article  sell  ? 

Note. — The  1st  discount  is  20?^  of  the  list  price,  and  the  2d  discount 
is  \0%  of  80^  of  list  price. 

135.  Commission. — Very  often  merchandise,  especially 
such  as  produce  and  live  stock,  is  bought  and  sold  in  distant 
markets.  Usually,  such  buying  and  selling  are  not  done  by  the 
parties  furnishing  the  money  and  the  commodities,  but  by 
agents  employed  for  that  purpose.  An  agent,  or  commission 
merchant,  charges  for  his  services  a  certain  per  cent  on  the 
amount  he  receives  in  selling  or  the  amount  he  pays  out  in 
buying.     This  charge  is  called  Commission. 

Principles:  1.  The  commission  is  always  some  number  of  per 
rent  of  the  price  of  that  which  is  bought  or  sold. 

2.   The  proceeds  are  equal  to  the  selling  price  minus  the  commission . 
S.   The  amount  is  equal  to  the  selling  price  plus  the  commission. 

TERMS. 

S,  price  of  the  sale  or  purchase. 

p,  number  of  per  cent  of  commission. 

0,  commission. 

P,  part  left  after  deducting  the  commission. 

A,  amount  of  investment  and  commission. 

Note. —  S  is  selected  to  represent  the  price  of  both  purchases  and 
mles  —  that  on  which  commission  is  based. 

Development  of  the  formulas  for  commission. 

C  =  ^ — .    (Prin.  1.) 
100 


262  ADVANCED   ARITHMETIC. 

Relation  I.  Abstractly,  the  commission  is  equal  to  the  product 
of  the  purchase  {or  sale)  and  the  number  of  per  cent  of  commission  j 
divided  by  100. 

To  find  the  proceeds :    (  Prin.  2.) 

100  100       ' 

p^s(im-p) 

100      • 

Relation  II.  Abstractly,  the  proceeds  are  equal  to  the  product  of 
the  purchase  (or  sale)  and  100  minus  the  number  of  per  cent  of 
commission,  divided  by  100. 

To  find  the  amount :    ( Prin.  3.) 

A  =  S  +  C  =  S+^',  or, 

100 

Relation  III.  Abstractly,  the  amount  of  investment  and  com- 
mission is  equal  to  the  product  of  the  purchase  (or  sale)  and  100 
plus  the  number  of  per  cent,  divided  by  100. 

examples. 

1.  1  sell  goods  for  $1430,  commission  4%.  Find  my  com- 
mission. 

„  -  ,.        ^    1430x4    __  _ 

Solution:  C  = =  57.2. 

100 

Answer,  $57.20. 

2.  How  much  must  an  agent  remit  to  his  principal,  if  he  sell 
goods  to  the  amount  of  $1200,  commission  2%  ? 

Solution:  P= =  1176. 

100 

Answer,  $1176. 


ADVANCED   ARITHMETIC. 


263 


3.  I  send  my  agent  $1442  to  be  invested  in  goods,  commis- 
sion 3%.     What  is  the  price  of  the  goods  bought? 

.^(100+p). 
100 
SxlOS 


Solution:  (1)  A 

(2)  1442  = 


100 


1442X100^^^^^ 
103 
Answer,  $1400. 

4..  A  lawyer  charges  $35  for  collecting  a  debt  of  $700.     Find 
his  %  of  commission. 


Solution :  (1) 

(2)p  = 


100 

35x100 


700 
Answer,  5^ 


=  5. 


EXERCISE  CXII. 


1.  What  is  commission  ?     A  commission  merchant  ? 

2.  Name  and  explain  the  terms  used  in  commission. 

3.  Develop  the  formulas  for  commission. 

4-.  Commit  to  memory  the  principles  and  relations. 


No. 

$S 

Pfo 

$c 

$P 

$A 

6. 

724 

50 

4 

? 

? 

6. 

840 

2^ 

? 

? 

7. 

920 

? 

? 

892 

40 

8. 

128 

80 

? 

6 

44 

? 

9. 

? 

? 

48 

75 

1901 

25 

10. 

? 

3 

? 

1225 

11 

11. 

? 

2 

26 

50 

? 

12. 

? 

H 

9 

588 

70 

13. 

? 

? 

14 

854 

264  ADVANCED    ARITHMETIC. 

i^.  A  commissiou  merchant  sells  8000  lb.  of  bacon  at  10/ 
per  pound ;  he  pays  freight  $50,  and  drayage  $2.  If  his  com- 
mission is  2%,  what  are  the  net  proceeds  ? 

Note. — "Net  proceeds"  are  the  balance  after  deducting  all  expenses. 

15.  An  agent  receives  $475  with  which  to  purchase  goods.  If 
his  commission  is  1^%,  what  amount  does  he  invest? 

16.  My  agent  sells  50  head  of  hogs  averaging  225  pounds,  at 
5/ per  pound.  He  invests  the  net  proceeds  in  wheat  at  $.50 
per  bushel.  His  commission  for  selling  is  8%  ;  for  buying, 
2% .     How  much  wheat  does  he  buy  ? 

17.  My  commission  at  2%  for  selling  600  hogs,  averaging  225 
lb.  each,  is  $135.  What  did  I  get  for  the  hogs  ?  how  much  per 
hog  ?  how  much  per  pound  ? 

18.  I  received  an  account  sales,  stating  that  my  net  proceeds 
were  $4301.80;  freight,  $55.70;  drayage,  $8.00;  commission, 
$135.     Find  the  rate  of  commission. 

Note. — An  ''account  sales''  is  a  final  statement  made  by  the  agent  to 
his  principal.  It  shows  (1)  the  selling  price  of  the  goods,  (2)  all  the  ex- 
penses, including  commission,  and  (3)  the  proceeds  to  be  remitted. 

19.  An  agent,  after  deducting  his  commission,  5%,  other 
expenses,  $42.24,  had  $988.76  to  remit.  Find  the  amount  of 
sales. 

20.  Sold  flour  at  3^%  commission  ;  invested  |  of  the  proceeds 
in  coffee  at  1^%  commission,  and  remitted  the  balance,  $482.50. 
What  was  the  value  of  the  flour,  the  coffee,  and  each  commis- 
sion ? 


The  commissions  are  taken  out  of  the  remaining  I  of  the  sales  of  the 
flour. 

136.   Stocks  and  Bonds. 

Note. — Children  in  the  public  schools  usually  know  little  or  nothing 
about  Stock  Companies  or  Corporations.    A  day  or  two  spent  in  working 


ADVANCED   ARITHMETIC.  265 

out  the  eight  examples  below  will  be  of  great  service  to  the  pupils.  Or- 
ganize the  Galena  Mining  Company,  have  pupils  represent  Jones,  Smith, 
Brown,  etc.,  write  out  some  of  the  certificates  of  stock,  discuss  the  mean- 
ing of  dividends,  surplus,  and  so  on.  Pupils  cannot  intelligently  solve 
problems  which  they  do  not  understand. 

CEKTIFICATE   OF  STOCK. 


No.    I.  Capital  Stock,  $200,000.    Shares,  $100  each.      £00  Shares. 

GALENA  MINING  COMPANY, 

GALENA.  KANSAS. 

This  !s  to  certify  that  [AyJUiuJuira.  J&yrJMx^  is  entitled  to 
c3u>o  (hL\J^'Y\xL\>ixL  Shares  in  the  Capital  Stock  of  the  Galena 
Mining  Connpany.  Transferable  only  on  the  books  of  the  Connpany, 
in  person  or  by  attorney,  upon  surrender  of  this  certificate. 

In  Witness  Whereof,  the  seal  of  said  Company  is 
(L.  S.)         hereto  affixed,  this   I  Otfv  day  of  9eiv.,    i8C|0. 
Attested  by  Cuxburu  OtAA/\A^, 

Secretary. 


The  Par  Value,  or  face  value,  of  a  share  of  stock  is  what  it  is 
valued  at  on  the  face  of  the  certificate. 

The  Market  Value  of  a  share  of  stock  is  its  selling  price  in  the 
market. 

Note. — When  a  share  sells  for  more  than  its  face  value,  it  is  said  to 
be  nhove  par,  or  at  a  premium.  When  a  share  sells  for  less  than  its  face 
value,  it  is  said  to  be  below  par,  or  at  a  discount. 

Stock  companies  usually  divide  their  earnings  (profits  or 
gains)  annually.  The  amount  thus  divided  is  called  the  divi- 
dend. Whoever  owns  a  share  of  stock  receives  the  dividend 
paid  on  that  share. 

Buying  or  selling  of  stocks  is  sometimes  done  through  a 
broker  (agent),  who  charges  for  his  services  a  brokerage  (com- 
mission) of  so  many  %  of  the  par  value  of  the  stocks  bought  or 
sold. 


266  ADVANCED    ARITHMETIC. 


EXAMPI.E8. 


The  Galena  Mining  Company  was  organized  and  incorporated 
under  the  laws  of  Kansas,  January  1,  1890,  with  the  following 
stockholders : 

William  Jones,  200  shares $20000 

William  Smith,  400  shares 40000 

James  Brown,  850  shares 35000 

Galen  Terry,  400  shares 40000 

G.  W.  Jolley,  650  shares 65000 

Capital  stock  (C.  S.) .$200000 

At  the  close  of  the  year  1890,  the  net  earnings  of  the  com- 
pany were  $22480.50. 

1.  If  the  company  reserves  a  surplus  in  the  treasury  of 
$2480.50,  what  will  be  the  rate  of  dividend  (R.  D.)? 

Note. —  After  reserving  a  surplus  to  pay  running  expenses,  what  is  left 
of  the  earnings  or  gains  is  divided  among  the  stockholders; 

Solution:   (1)  $22480.50 -$2480.50  =  $20000.     Amt.  to  be  divided. 

(2)  $20000  =  (  )%  of  C.  S.  ?    (  Question.) 

(3)  $200000  =  100^  of  O.S.    (Basis.) 
tV  of  (3)  =  (4)  $20000=10^  of  0.  S.,  answer. 

2.  What  is  Smith's  share  of  the  dividend  (his  C.  S.,  $40,000)? 

Solution  :  (1)  10^  of  his  0.  S.  =  $(  )  ?    (Question  ) 
(2)  100^  of  his  0.  S.  =  $40000.     (Basis.) 
iV  of  (2)  =  (3)  10^  of  his  C.  S.  =  $4000,  answer. 

Note. — Brokerage  is  the  commission  an  agent  (broker)  charges  for 
buying  or  selling  stocks  or  bonds.  But  unlike  commission,  brokerage  is 
always  some  per  cent  of  the  Par  Value  of  the  stocks  or  boTK^s,  and  not  some 
per  cent  of  the  cost. 

3.  Jones  sells  some  of  his  shares  to  W.  W.  Miller  at  90  (10% 
below  par),  brokerage  ^  %,  for  $13518.75.  Find  the  face  value 
of  the  stocks  sold. 


ADVANCED    ARITHMETIC.  267 

Solution  :   (1)  90^  of  par  value  =  price  of  stocks. 

(2)  I ^  of  par  value  =  brokerage. 

(3)  100^  of  par  value  =  $(  )?    (Question.) 

(4)  m%  of  par  value  =  $13518.75.     (Basis.) 
ylT  of  (4)  =  (5)  1%  of  par  value  =  $150. 

100  X  (5)  =  (6)  100^  of  par  value  =  $15000,  answer. 

At  the  beginning  of  the  year  1891,  Terry  sells  his  shares  to 
J.  O.  Bowers;  Jolley  sells  300  shares  to  G.  W.  Greene  at  120 
(20%  above  par) ;  at  the  end, of  the  year  1891,  the  company 
finds  that,  after  reserving  a  surplus  of  $3124.40,  it  can  declare 
a  dividend  of  6%. 

Note. — For  the  sake  of  brevity  the  ^'question  "  will  be  omitted  in  the 
remainder  of  these  solutions. 

4-.  Find  the  net  earnings  of  the  company  for  1891. 

Solution  :   (1)  100^  of  C.  S.  =  $200000.     (Basis.) 
T^  of  (1)  =  (2)  1^  of  0.  S.  =  $2000. 
6x(2)  =  (3)  6^of  0.  S.  =  $1^000. 

(4)  $12000+3124.40.  =$15124.40,  answer. 

5.  What  did  Greene's  800  shares  cost  him  (at  120)  ? 

Solution:   (1)  100^  of  300  shares  =  $30000.     (Basis.) 
T^jy  of  (1)  =  (2)  1%  of  300  shares  =  $300. 
120  X  (2)  =  (3)  120^  of  300  shares  =  $36000,  answer. 

6.  What  was  Greene's  dividend  ? 

SoZw^ion  :  (1)  100^  of  300  shares  =  $30000.  (Basis.) 
T^^  of  (1)  =  (2)  1%  of  300  shares  =  $300. 

6  X  (2)  =  (3)  Q%  of  300  shares  =  $1800,  answer. 

7.  What  is  Greene's  rate  of  income  on  his  investment  ? 

Note. — The  *'rate  of  income  "  is  always  some  number  of  per  cent  of  the 
investment  or  cost. 

Solution  :   (1)  $36000  =  100^  of  his  investment.     (Basis.) 
aV  of  (1)  =  (2)  $1800  =  5^  of  his  investment,  answer. 


268  ADVANCED    ARITHMETIC. 

8.  If  Bowers 's  rate  of  income  is  8%  for  1891,  for  how  much 
below  par  did  he  buy  his  stock  ?  (R.  D.  was  6%). 

Parts  :   (a)  8^  of  investment  =  6^  of  his  C.  S. 

(6)  IW  of  investment  =  (  )%  of  his  0.  S.? 
Solution  :  (1)  8^  of  investment  =  6^  of  C.  S. 
I  of  (1)  =  (2)  1%  of  investment  =  1%  of  0.  S. 
•      100  X  (2)  =  (3)  100^  of  investment  =  75^  of  0.  S.,  or  2b%  below 
par,  answer. 

Bonds  are  promissory  notes  issued  by  a  public  corporation 
(nation,  state,  county,  city),  or  a  private  corporation  (railroad 
company,  banking  company,  mining  company,  etc.).  Bonds 
have  a  face  value  (par  value),  a  rate  of  interest,  and  when  bought 
and  sold  they  have  a  market  value.  Brokerage  may  be  charged 
by  an  agent  for  buying  or  selling  bonds. 

EXAlCPIiES. 

1.  I  own  6's  to  the  amount  of  $5500;  what  is  my  annual  in- 
come, or  interest  ? 

Note. — **  6's  "  means  that  the  rate  of  interest  is  Q%. 
Solution  :  (1)  100^  of  P.  V.  =  $5500.  (Basis.) 
i^^of  (1)  =  (2)  l^of  P.  V.  =  $55. 

6  X  (2)  =  (3)  6^  of  P.  V.  =  $330,  answer. 

2.  A  buys  bonds  amounting  to  $50000  at  104,  brokerage  \%  ; 
what  do  they  cost  him  ? 

Solution  :   (1)  100^  of  P.  V.  =  $50000.  (Basis.) 

^U  of  (1)  =  (2)  1%  of  P.  V,  =  $500. 

104i  X  (2)  =  (3)  104i^  of  P.  V.  =  $52125,  answer. 

3.  A  invests  $7210  in  bonds  at  90,  brokerage  ^%  ;  what  is  the 
face  value  of  the  bonds  ? 

Solution  :   (1)  901^  of  P.  V.  =  $7210.     (Basis.) 
yfy  of  (1)  =  (2)  1^  of  p.  V.  =  $80. 
100  X  (2)  =  (3)  100^  of  P.  V.  =  $8000,  answer. 


ADVANCED    ARITHMETIC. 


269 


!^.  B  invests  $24000  in  4's,  at  80;   what  is  his  interest  ? 

Solution  :   (1)  80^  of  P.  V.  =$24000.     (Basis.) 
i^  of  (1)  =  (2)  1^  of  P.  V.  =  $300. 
4  X  (2)  =  (3)  4%  of  P.  V.  =  $  1200,  answer. 

5.  How  must  C  buy  G's  to  yield  an  income  of  5%  ? 

Solution  :   (!)  5%  of  M.  V.  =  Q%  of  P.  V.     (Basis.) 
20x(l)  =  (2)  lOO^of  M.  V.  =  120^of  P.  V.,answer. 

6.  Which  furnishes  the  better  income  on  the  investment — 
5's  at  75,  or  6's  at  80  ? 

In  the  First :   (1)  75^  of  P.  V.  =  100^  of  investment.     (Basis.) 
tV  of  (1)  =  (2)  6%  of  P.  V.  =  6|^  of  investment. 

In  the  Second :   (1)  80^  of  P.  V.  =  100^  of  investment.     (Basis.) 
^  of  (1)  =  (2)  1%  of  P.  V.  =  1%  of  investment. 
6  X  (2)  =  (3)  Q%  of  P.  V.  =  7i^  of  investment. 

(4)  71%  of  investment -61^  of  investment  =  |^  of 
investment,  in  favor  of  the  second,  answer. 

The  problems  of  stocks  and  bonds  are  so  nearly  alike  that 
one  set  of  principles  and  formulas  applies  to  both. 

Principles:    1.   The  discount,  premium ^  dividend  (or  interest) 
and  brokerage  are  each  some  number  of  per  cent  of  the  par  value. 

2,  The  per  cent  {or  rate)  of  income  is  some  number  of  per  cent 
of  the  cost. 

3.  The  market  value  is  equal  to  the  par  value  plus  the  premium, 
or  minus  the  discount. 

If.   The  cost  is  equal  to  the  market  value  plus  the  brokerage. 

terms. 
P,  par  value. 
M,  market  value. 

C,  cost. 

B,  brokerage. 

D,  dividend  (stocks). 
/,  interest  (bonds). 


270  ADVANCED    ARITHMETIC. 

In.,  income. 

Pr.,  premium. 

Dis.,  discount. 

R,  number  of  %  of  the  dividend  or  interest. 

r,  number  of  %  of  the  income  on  the  investment. 

j9,  number  of  %  of  premiupi  or  discount. 

_p',  number  of  %  of  the  brokerage. 

Develop  the  formulas  for  Stocks  and  Bonds  : 

pP 
1.  Pr.  or  Dis.=f-— .    (Prin.  1.) 

^'       100'  ^^'■'''-  ^'^ 

T.  r        "fP 

3.  D  or  I=-Yqq'    (Prin.  D 

Since   the   income  on  the  investment  is  the  same  as   the  dividend 
(stocks)  or  the  interest  (bonds),  it  follows  that — 

rC 

4.  D  or  1^1^.    (Prin.  2.) 

When  there  is  no  brokerage, 

5.  D  or  /=^.    (Why?) 

By  comparing  formulas  3  and  4,  we  obtain 
RP    rC 
100"  100'  °^' 

6.  RP=rC. 
When  there  is  no  brokerage, 

7.  RP=rM. 

When  stocks  or  bonds  are  at  a  premium, 

(1)  M=P+Pr.     (Prin.  3.) 

(2)  C  =  P4-Pr+5.     (Prin.  4.) 


ADVANCED   ARITHMETIC.  271 

P(100+p) 
8.  M-         ^^         . 

(2) -(4)  c  r\P^\P'^   mp+pp+p'p 

(2)-(4)  C-^+jgg+^^-  ^^  ,or, 

P(100+p+/) 

-^^  ^~        loo       • 

When  stocks  or  bonds  are  at  a  discount,  show  that — 

EXAMPLES. 

Note. — In  this  book,  the  par  value  of  a  share  or  a  bond  is  $100,  unless 
it  should  be  otherwise  stated. 

1.  Find  the  discount  on  20  shares  of  stock  sold  at  20%  be- 
low par. 

Solution:  Dis.  =  — — r —  =400. 
100 

Answer,  $400, 

2.  Find  the  premium  on  60  shares  of  stock  sold  at  12^% 

premium. 

^  ,  ^.         „       12^x6000    ^^^ 

Solution  :  Pr.  =  — ^~ =  750. 

100 

Answer,  $750. 

3.  Find  the  brokerage  at  i%  on  40  shares  of  stock. 

c,  ,  ,.         „     ix4000    ^ 
Solution:  ^  =  -^-——-  =  5. 

Answer,  $5. 


4^.  Find  the  dividend  on  50  shares  of  8%  stock. 
^     8x5000    , 
^  =  -10^  =  ' 
Answer,  $400. 


Solution:  D  =  l^i^  =  400. 


272  ADVANCED    ARITHMETIC. 

5.  Find  the  rate  of  interest  on  50  bonds  that  yield  $450  an- 
nual interest. 

Solution:  (1)  I=^r^. 

(2)450  =  ^. 

Answer,  %. 

6.  If  the  annual  interest  of  $920  on  bonds  pays  an  income  of 

8%,  find  the  cost. 

rC 
Solution:  (1)  -^=77^- 

(2)  920  =  ||. 

(3)  C  =  8?^  =  11500. 

O 

Answer,  $11500. " 

7.  Find  the  cost  of  45  shares  of  stock  bought  at  110. 

^  .  ^.         -.    4500x110     .__ 
Solution:  M= r— =  4950. 

J-UU 

Answer,  $4950. 
Note.— The  "110"  =  (100+p)  =  (100  +  10). 

8.  Find  the  cost  of  15  bonds  at  60,  brokerage  -|. 

Solution:  (7  =  -^^^^^  =  901.875. 
Answer,  $901,875. 
Note.— The  60i  =  (100 -p+pO  =  100- 40+i). 

9.  I  buy  6%  bonds  at  80;  find  my  rate  of  income. 
Note.— Assuming  1  bond  as  a  basis,  the  P  is  $100,  the  M  is  $80. 

Solution:  (1)  RP=rM. 


(2)  6xlOO  =  80r. 
,3)  rJ-^^-V. 
Answer,  11%. 


ADVANCED   ARITHMETIC.  278 

10.  I  bought  5's,  ^  brokerage,  for  $903.75.  If  they  pay  me 
^2TT%  Oil  my  investment,  find  the  par  value,  the  %  of  dis- 
count, the  brokerage,  and  the  market  value. 

Note. — ''  5's  "  means  bonds  that  bear  b%  interest  per  annum. 

Solution:  (1)  RP=rC, 

(2)  5  P=8/A:X  903.75. 

(3)  p^2000x  903.75^ 

241x5 
Par  value,  $1500. 

Brokerage,  $3.75. 

(5)  if=  $903.75 -$3.75  =  $900. 

(6)  Dis.  =  $1500 -$900  =  $600. 

Discount,  40^. 
EXERCISE  CXIII. 

1.  Discuss  the  organization  of  a  stock  company  —  the  capital 
stock,  the  issuing  of  shares,  and  the  distribution  of  earnings. 

2.  Define  brokerage,  par  value,  market  value,  premium,  dis- 
count. 

3.  When  a  company  loses  instead  of  gaining  during  the  year, 
how  is  the  assessment  made  to  pay  this  loss  ? 

J^.  What  is  a  bond  ?     The  interest  ? 

5.  Commit  to  memory  the  principles. 

6.  Develop  the  formulas,  and  write  out  the  relation  expressed 
by  each  formula. 

7.  Find  the  discount  on  34  shares  of  stock,  sold  in  market  at 
30%  below  par. 


274  ADVANCED    ARITHMETIC. 

8.  Find  the  premium  on  75  shares  of  10%  stock,  sold  in  mar- 
ket at  30%  premium. 

9.  I  bought  bonds  at  105  for  $7350.     Find  the  par  value. 

10.  40  shares  of  stocks  sold  in  market  for  $3600.  Find  the 
per  cent,  of  discount. 

11.  I  bought  80  shares  of  6%  stock.  Find  my  annual  divi- 
dend. 

12.  My  interest  on  6%  bonds  is  $540.  Find  the  face  value 
of  the  bonds. 

13.  I  receive  a  dividend  of  $400,  which  is  6|%  on  my  invest- 
ment.    Find  the  cost  of  the  bonds. 

14..  I  bought  60  shares  of  8%  stock  for  110.  Find  the  rate  of 
income. 

15.  I  bought  5%  stock  at  70.     Find  the  rate  of  income. 

16.  I  bought  6%  bonds  so  as  to  pay  me  10%  on  my  invest- 
ment.    Find  the  market  price. 

17.  Find  the  cost  of  40  shares  of  stock,  sold  at  80,  brokerage 
i%. 

18.  Find  the  cost  of  75  shares  of  stock,  sold  at  120,  broker- 
age i%. 

19.  How  many  shares  of  stock  at  77 J,  brokerage  i%,  can  be 
bought  for  $15600  ? 

20.  Which  pays  the  better  income  on  the  investment,  and 
how  much  —  7's  at  90,  or  6's  at  75  ? 

21.  A  man  owns  5%  railroad  stocks  worth  90  and  4%  bonds 
worth  80.  If  his  income  from  each  is  $720,  find  the  par  value 
of  each. 

22.  In  No.  21,  find  the  rate  of  income  on  each  investment. 

23.  How  much  money  must  be  invested  in  4's  at  90  to  yield 
an  annual  income  of  $120? 

21/..  On  bonds  bought  at  70,  my  rate  of  income  is  5%.  This 
income  annually  amounts  to  $1750.  Find  the  amount  of  the 
bonds  and  the  rate  of  interest. 


ADVANCED    ARITHMETIC.  275 

137.  Taxes. —  A  tax  is  a  sum  of  money  levied  on  persons 
(poll  tax)  or  property  (property  tax). 

Principles:  1,  A  poll  tax  is  a  certain  sum  required  usually  of 
those  citizens  who  have  the  right  to  vote. 

2.  A  property  tax  is  some  number  of  per  cent  of  the  value  of  the 
property  taxed. 

TERMS. 

F,  value  of  property. 

p,  number  of  %  of  the  taxes. 

r,  taxes. 

Formula:   T=^.     (Prin.  2.) 

EXAMFIiEB. 

1.  In  a  certain  county  there  were  7570  polls,  at  $1.25.  Find 
the  poll  tax. 

Solution:  r=  7570  x  $1.25  =  .$9462.50,  answer. 

2.  The  property  of  the  county  was  valued  at  $5897500.  Find 
the  tax  raised  by  a  i%  (5  mills)  levy. 

Solution:  r=  ^^^^^^^  =  29487.50. 
100 

Answer,  $29487.50. 

Note. — In  tax  levies  is  almost  the  only  place  in  business  where  the 
denomination  mill  is  used.  1  mill  =  i^,<*.  A  5-mill  levy  is  a  levy  of  5 
mills  or  J^  on  $1 ;  or,  |^  of  the  value  of  the  property  taxed. 

EXERCISE  CXIV. 

1.  Define  poll  tax;  property  tax. 

2.  Give  the  principle  governing  poll  tax ;  property  tax. 

3.  Give  the  relation  expressed  by  the  formula  above. 

4..  If  the  number  of  polls  be  represented  by  N,  and  the  tax 
on  each  poll  by  t,  give  the  formula  for  finding  poll  tax,  T. 
5.  Find  the  tax  on  1200  polls,  at  $1.85  each. 


276  ADVANCED    ARITHMETIC. 

6.  Find  the  property  tax  oii  a  farm  valued  at  $10400,  if  the 
levy  is  14  mills  (lf%). 

7.  What  is  the  valuation  of  A's  property  who  pays  $51  on  a 
15  mill  m%)  levy? 

8.  A  school  district  has  $20000  taxable  property,  and  desires 
to  raise  $240  teachers'  wages.    What  %  levy  must  they  run  ? 

9.  A  tax  of  $787.50  is  raised  on  25  polls  at  $1.50  each,  and  a 
property  tax  of  6  mills.    Find  the  value  of  the  property  taxed. 

10.  Find  the  %  of  the  tax  levy,  when  $17500  worth  of  prop- 
erty, and  40  polls  at  $1  each,  produce  a  tax  of  $285. 

138.  Duties. — Taxes  levied  by  the  government  upon  im- 
ported merchandise  are  called  Duties.  Duties  are  of  two 
kinds,  specific  duties  and  ad  valorem  duties. 

A  Specific  duty  is  levied  upon  the  amount  of  the  merchandise 
imported.  An  Ad  valorem  duty  is  levied  upon  the  value  of  the 
merchandise  imported. 

Tare,  leakage,  and  breakage  are  allowances  deducted  from  the 
gross  amount  of  imported  merchandise.  These  deductions  are 
made  before  levying  duties. 

Tare  is  the  deduction  made  for  the  weight  of  the  boxes,  bar- 
rels, etc.,  containing  the  merchandise. 

Leakage  is  the  deduction  made  for  the  loss  of  liquids  shipped 
in  barrels  or  casks.  The  amount  to  be  deducted  is  determined 
by  gauging. 

Breakage  is  the  deduction  made  for  the  loss  of  liquids  shipped 
in  bottles. 

Principles:  1.  Specific  duty  is  a  certain  sum  per  pound,  yard, 
gallon,  etc.,  on  the  merchandise  imported. 

2.  Ad  valorem  duty  is  some  number  of  per  cent  of  the  net  value 
of  the  merchandise  imported. 

TERMS. 

V,  net  value  of  merchandise. 

p,  number  of  %  of  ad  valorem  duty. 

D,  the  duty. 


ADVANCED    ARITHMETIC.  277 

Formula  for  ad  valorem  duty  : 

vV 
D=-^—-.     (Prin.  2.) 
100 

EXAKPIiES. 

1.  What  is  the  duty  on  560  bottles  of  wine,  duty  60^  per 
dozen,  breakage  12^%  ? 

Solution :  (1)  Breakage  =  12J^  of  560  bottles  =  70  bottles. 

(2)  560  bottles -70  bottles  =  490  bottles  =  40|  dozen  bottles. 

(3)  i)M«i/  =  40|x  60^  =  $24.50,  answer. 

2.  Find  the  duty  on  400  bags  of  coffee,  each  weighing  180  lb. 
gross,  invoiced  at  8/  per  lb.,  tare  5%,  ad  valorem  duty  20%. 

Solution  :  (1)  Amount  =  400  x  180  lb.  =  72000  lb. 

(2)  Tare  =  b%  of  72000  lb.  =  3600  lb. 

(3)  72000  lb.  -  3600  lb.  =  68400  lb. 

(4)  raZMe  =  68400  x8<*  =  $5472. 

20X5472^ 
100 
Answer,  $1094.40. 

3.  A  merchant  imports  10  casks  of  brandy,  each  containing 
80  gallons,  worth  $1.40  per  gallon;  duties  are  10/  per  gallon, 
and  25%  ad  valorem;  leakage,  10%.     Find  the  whole  duty. 

Solution  :  (1)  Amount  =  10  X 30  gal.  =  300  gal. 

(2)  Leakage  =  10^  of  300  gal.  =  30  gal. 

(3)  300  gal.  -30  gal.  =  270  gal. 

(4)  Value  =  270  X  $1.40  =  $378. 

(5)  Sp.  2).  =  270xlO,<^  =  .$27. 

(6)  Ad.  D.  =  25^  of  $378  =  $94.50. 

(7)  $27 +$94.50  =  $121.50,  answer. 

EXERCISE  CXV. 

1.  Define  duty,  specific  duty,  ad  valorem  duty,  tare,  leakage, 
breakage. 

2.  Commit  the  principles. 

8.  Give  the  relation  expressed  by  the  formula. 


278  ADVANCED   ARITHMETIC. 

Jf,  Write  the  formula  for  specific  duty  (i)),  when  the  amount 
(^)  of  the  merchandise  and  the  duty  (cZ)  on  the  unit  of  amount 
is  given. 

5.  Find  the  duty  on  500  yd.  of  silk  invoiced  at  $1.50,  at  20% 
ad  valorem. 

6.  What  is  the  duty  at  25%  ad  valorem  on  400  boxes  of 
raisins  of  40  lb.  each,  invoiced  at  9/  per  lb.,  tare  10%  ? 

7.  The  ad  valorem  duty  on  1500  yd.  Irish  linen,  valued  at 
40/  per  yd.,  is  $210.     Find  the  %  of  duty. 

8.  Find  the  specific  duty  on  750  yd.  broadcloth  at  50/  per  yd. 
P.  What  is  the  specific  duty  per  gallon,  when  a  duty  of  $540 

is  paid  on  50  casks  of  wine  of  30  gallons  each,  leakage  10%  ? 

10.  Find  the  whole  duty  on  4500  lbs.  of  wool,  invoiced  at  20/ 
per  lb.,  specific  duty  10/  per  lb.,  ad  valorem  duty  12%,  tare 


139.  Insurance. —  Insurance  is  indemnity  against  loss 
by  damage  or  death.  Insurance  is  of  two  kinds,  2Jroperty  in- 
surance and  personal  insurance. 

Property  insurance  is  indemnity  for  loss  of  property,  caused 
by  fire,  lightning,  tornadoes,  death  of  stock,  and  the  like. 

Personal  insurance  is  indemnity  for  loss  caused  by  death, 
sickness,  or  accident. 

A  Policy  is  the  written  contract  between  the  insurance  com- 
pany and  the  person  insured. 

A  policy  which  insures  the  payment  of  a  certain  sum  only  at 
the  death  of  the  person  insured,  is  called  a  Life  policy.  A  pol- 
icy which  insures  the  payment  of  a  certain  sum  at  a  specified 
time,  or  at  death  occurring  before  that  time,  is  called  an  En- 
doivment  policy. 

Premium  is  the  sum  paid  for  insurance. 

The  Risk  is  the  amount  insured. 

Note. — Insurance  companies  seldom  agree  to  insure  property  for  its 
full  value.    They  usually  make  the  risk  i  or  |  of  the  full  value. 


ADVANCED    ARITHMETIC.  279 

Principles:  1.  In  property  insurance,  the  pfemium,  due  an- 
nually, is  some  number  of  per  cent  of  the  risk. 

2.  In  life  insurance,  the  premium  is  usually  a  certain  sum  per 
^1000,  payable  annually  {semi-annually  or  quarterly) . 

TERMS. 

i?,  the  risk. 

p,  number  of  %  of  the  premium. 

P,  the  premium. 

Formula  for  Property  Insurance  : 

EXAMPLES. 

1.  Find  the  premium  at  2%  on  a  risk  of  $8400. 

Solution:   P=^^^~  =  m. 
Answer,  $168. 

2.  The  annual  premium  is  $480,  paid  on  a  factory  valued  at 
$48000,  insured  at  f  of  its  value.     Find  the  rate  of  insurance. 

Solution  :  (1)  |  of  $48000  =  $32000,  risk. 
(2)480  =  ^^^^50.    (Why?) 

._,         480x100     ,, 
^^>  P=     32000     '^^- 

Answer,  1|^.  , ' 

3.  A  company  insured  a  vessel  worth  $60000  for  f  its  value    ^    , 
at  2%  ;  they  reinsured  |  of  their  risk  at  If  %.     Find  the  pre-      .! 
mium  retained  by  the  first  company. 

Solution  :  (1)  f  of  $60000  =  $40000,  whole  risk. 

(2)  I  of  $40000  =  $30000,  second  risk. 

(3)  1st  premium  =  2%  of  $40000  =  $800. 

(4)  M  premium  =  1^^  of  $30000  =  $525. 

(5)  $800  -  $525  =  $275,  answer. 


ADVANCED    ARITHMETIC. 

If.  What  will   be   the   annual  premium  on  a  life  policy  ofj 
$4000  at  $25.40  per  $1000  ? 

Solution:  Premmw  =  4 x $^5.40  =  $101.80,  answer. 

o^.  If  Mr.  Jones  was  insured  for  $5000,  at  an  annual  pre- 
mium of  $86.40  per  $1000,  what  amount  did  the  company  pay 
out  more  than  it  received,  if  Mr.  Jones  died  after  making  11 
annual  payments  ? 

Solution:   (1)  PremfMm  =  11  x5x $36.40  =  $2002.     (Why?) 
(2)  $5000 -$2002  =  $2998,  answer. 

EXERCISE  CXVI. 

^•^jy^jiliUc.  1.  Define  insurance,  policy,  premium. 

2.  Define  a  life  policy,  endowment  policy. 

3.  Give  the  principles. 

4-.  Give  the  relation  expressed  by  the  formula. 
•      6.  How  do  you  find  the  premium  on  a  life  insurance  policy, 
when  the  amount  of  the  policy  and  the  premium  per  $1000  is 
given  ? 
f  \j,.)^6.  Find  the  premium  at  3%  on  a  risk  of  $6500. 
'  ',  v"*  7.  What  is  the  rate  of  insurance,  if  the  x:)remium  on  $5420  is 
,>  $135.50? 
^'\-s/''8.  The  premium  at  2}%  on  a  certain  risk  was  $99.     Find  the 
risk. 

9.  A  vessel  valued  at  $50400  was  insured  for  |  its  value  at 
1^%.  The  first  company  re-insured  |  of  the  risk  in  a  second 
company,  paying  $441  premiuHi.  Find  the  first  premium,  and 
the  second  rate  of  insurance. 

10.  Find  the  premium  paid  in  15  yr.  on  a  life  policy  of 
$6000,  at  $27.50  per  $1000  annually. 

11.  A  man  took  out  an  endowment  policy  for  $4000  and  died 
after  making  10  annual  payments.  If  his  estate  received  $2682 
more  than  he  had  paid  out,  what  was  the  annual  premium  per 
thousand  ? 


-V 


ADVANCED    ARITHMETIC.  281 

3.    PERCENTAGE  "WITH  TIME. 

140.  Terms  Used  in  Interest.    Interest  is  a  sum 

paid  for  the  use  of  money.  The  amount  of  money  used  is 
the  Principal.  The  per  cent  of  the  principal  which  equals 
one  year's  interest  is  the  Rate.  If  only  the  principal  bear 
interest,  the  interest  is  called  Simple  Interest. 

When  by  the  terms  of  a  note  interest  is  payable  annually, 
(semi-annually,  or  quarterly,)  and  is  not  paid  when  due,  this 
annual  interest  (interest  due  annually,  semi-annually,  or 
quarterly)  bears  interest  at  the  stipulated  or  legal  rate  until 
paid.  This  is  called  Annual  Interest.  A  note  which 
reads,  "and  if  the  interest  be  not  paid  annually  [semi-annu- 
ally, or  quarterly],  it  becomes  as  principal,  and  bears  the  same 
rate  of  interest,"  bears  Compound  Interest. 

141.  Time.  Since  the  rate  in  interest  problems  is  ^^per 
annum, ^^  or  by  the  year,  the  time  should  be  expressed  in  years. 
By  the  common  method  of  counting  time  in  interest  problems, 

lmo.=:yVyr. 
1  da.=^io  yr. 

This  method  of  counting  time  is  called  the  common  interest 
method. 

Example:  Express  in  terms  of  years  (1)  3  mo.,  (2)  17  da., 
(3)  6  mo.  8  da.,  (4)  5  mo.  18  da.,  (5)  2  yr.  4  mo.  12  da. 

Results  :  (1)  3  mo.  =  j\  yr. 

(2)  17da.  =  ^«Vyr. 

(3)  6  mo.  8  da.  =  iU  jr. 

(4)  5mo.  18da.  =  y^yr. 

(5)  2  y r.  4  mo.  12  da.  =  ^  yr. 
(6)'  1  yr.  2  mo.  7  da.  =  iU  yr. 

Note.— In  using  these  expressions  in  the  interest  problems  to  follow, 
cancellation  will  be  used  after  the  whole  problem  is  stated ;  therefore 


282  ADVANCED   ARITHMETIC. 

these  expressions  need  not  be  reduced  to  lowest  terms.  When  the  num- 
ber of  days  is  a  multiple  of  3,  divide  by  3  and  express  the  result  as  tenths 
of  a  month ;  as  in  (4)  and  (5)  above.    Thus : 

15da.  =  j^yr. 

33da.  =  ^yr. 

g 
24  da.  =^  yr.,  and  so  on. 

EXERCISE  CXVII. 

Express  in  years : 


1. 

6  mo. 

2. 

8  mo. 

S. 

9  mo. 

4. 

5  mo. 

5. 

10  mo. 

6. 

30  da. 

7. 

12  da. 

8. 

24  da. 

9. 

2  yr.  4  mo. 

10. 

1  yr.  8  mo. 

11. 

5  mo.  15  da. 

12. 

6  mo.  24  da. 

13. 

7  mo.  21  da. 

U- 

1  yr.  6  mo.  12  da. 

15. 

2  yr.  4  mo.  10  da. 

16. 

1  yr.  11  mo.  23  da 

Find  the  time  in  years  from  — 

17.  Mar.  20,  1900,  to  Sept.  25,  1901. 


yr. 

mo. 

da. 

1901 

9 

25 

1900 

3 

20 

Form : 

19C 

1         6  5  =  myr. 

Note. — Subtract  as  in  compound  numbers  (p.  176). 

i<5..Sept.  10,  1890,  to  Jan.  20,  1896. 

19.  June  1,  1897,  to  Jan.  5,  1898. 

20.  July  10,  1895,  to  June  5,  1896. 

21.  Feb.  1,  1899,  to  Nov.  1,  1900. 

22.  Dec.  25,  1884,  to  July  5,  1889. 

23.  Nov.  21,  1893,  to  Oct.  15,  1895. 
21^.  Aug.  30,  1898,  to  April  10,  1900. 


ADVANCED    ARITHMETIC.  288 

In  Bankers^  Interest  (bank  discount),  the  exact  number  of 
days  is  counted;  but,  in  expressing  the  time  in  years,  860  da. 
is  considered  1  yr. 

Example:  Find  the  time  in  years  from  Jan.  4,  1900,  to  Feb. 
15,  1900  (bankers'  interest). 

Process  :  (1)  From  Jan.  4, 1900,  to  Feb.  15, 1900  =  42  da. 
(2)  42da.  =  s»^yr. 

Note. — After  Jan.  4,  there  are  27  da.  in  Jan.  and  15  da.  in  Feb. ;  or, 
42  da.  in  all. 

In  Exact  Interest,  the  exact  number  of  days  is  counted;  but, 
in  expressing  the  time  in  years,  865  days  is  considered  1  yr. 
(866  days  in  leap  year). 

Example  :  Find  the  time  in  years  from  July  9,  1897,  to  Oct. 
12,  1897  (exact  interest). 

Process  :  (!)  From  July  9,  to  Oct.  12  =  95  days. 
(2)  95da.  =  /^yr. 

EXERCISE  CXVIII. 

Find  the  time  in  years  {bankers^  interest)  from  — 

1.  Feb.  12,  1900,  to  July  1,  1900. 

2.  Feb.  4,  1899,  to  April  10,  1899. 
8.  Mar.  17,  1896,  to  June  25,  1896. 

4.  Jan.  19,  1896,  to  May  10,  1896. 

5.  July  29,  1895,  to  Oct.  11,  1895. 

6.  Dec.  10,  1898,  to  Jan.  27,  1899. 

Find  the  time  in  years  (exact  interest)  from — 

7.  Nov.  12,  1898,  to  Jan.  1,  1894. 

8.  Oct.  21,  1900,  to  Dec.  21,  1900. 

9.  May  5,  1896,  to  Aug.  1,  1896. 

10.  Sept.  28,  1892,  to  Dec.  1,  1892. 

11.  Jan.  12,  1896,  to  Mar.  1,  1896. 

12.  Dec.  20,  1897,  to  Feb.  15,  1898. 


284  ADVANCED   ARITHMETIC. 

142.   Simple  Interest.     (See  definition,  p.  281.) 

TERMS. 

P,  the  principal. 

T,  the  time. 

R,  the  number  of  %  of  the  rate. 

/,    the  interest. 

A,  the  amount  of  principal  and  interest. 

Develop  the  formulas  for  simple  interest. 

We  know — 

(1)  Int.  on  $1  for  1  yr.  at  1^  =  $t^. 

(2)  Int.  on  $  P  for  1  yr.  at  1^  =  $  ^ 


(3)  Int.  on  $Pfor  Tyr.  at  !%■■ 


100 
PxT 
100  * 
PxTxR 


(4)  Int.  on  $P  for  T  yr.  at  R%=  . 

Now,  expressing  (4)  abstractly,  we  have  the  interest  formula- 

PRT 


1.   I: 


100  • 


Relation  I.  Abstractly,  the  interest  is  equal  to  the   continued 
product  of  the  principal,  rate,  and  time,  divided  by  100. 


We  know — 

PRT 


(1)  ^  =  P+7=P+- 


100  ' 


.,,     .^,    .     100  P+Pi?r 

^1^=^'^^= ^0 '^^' 

2.  ^^/(lOO+PT) 


100 

Relation  II.  Abstractly,  the  amount  of  a  given  principal  bear- 
ing interest  at  a  given  rate  for  a  given  time  is  equal  to  the  product 


ADVANCED    ARITHMETIC. 


285 


of  the  principal  and  100  plus  the  product  of  the  number  of  per  cent 
of  the  interest  and  the  time,  divided  by  100. 

Note.— Remember  that  in  using  these  formulas  the  time  must  be  ex- 
pressed in  years  before  applying  the  formula. 

EXAiyiPIiES. 

1.  Find  the  interest  on  $800  for  2  yr.  6  mo.  at  8%. 

Solution  :  (1)  2  yr.  6  mo.  =2|  yr.  =  |  yr. 
300x8x5 
^^^^-    100x2    "^^• 
Answer,  $60. 

2.  Find  the  interest  on  $1000  at  8%  for  1  yr.  3  mo.  20  da. 

Solution  :  (1)  1  yr.  3  mo.  20  da=  tH  yr. 
1000x8x470,  Q 
^"^^  ^        100x360     -l^-^^- 
Answer,  $104.44^. 

3.  At  what  rate  will  $800  gain  $86  in  1  yr.  4  mo.  ? 

Solution  :  (1)  1  yr.  4  mo.  =  i  yr. 

(2)/  =  ^^^ 


(3) 


100 
300x4xie 
3x100 


...    ^     36x3x100     - 
^^^^=     300x4     =^' 
Answer,  9%. 


Jf..  In  what  time  will  $400  gain  $87.60,  interest  at 

400x8xr 


Solution:  (1)  37.6  = 


100 


37.62000 
^^^  ^"400x8    -^•^^^• 

Answer,  1.175  yr.  =  1  yr.  2  mo.  3  da. 

5.  What  principal  will  gain  $39  in  1  yr.  3  mo.  at 

Solution  :  (1)  1  yr.  3  mo.  =  f  yr. 

(2)39  =  «^^><^ 


(3)P 


4x100 
39x4x100 
6x5 
Answer,  $520. 


=  520. 


^,  K  .-<  Y 


i  ^ 


ci^V^ 


Of 


286  ADVANCED    ARITHMETIC. 

6.  Find  the  amount  of  $640  at  9%  interest  for  1  yr.  5  mo. 

15  da. 

17  5 
Solution  :  (1)  1  yr.  5  mo.  15  da.  =  — -^yr. 

640x9xl7.5_ 
^"^^  ^         100x12      "^^• 

Interest,  $84. 
(3)  Amount  =  $640 + $84  =  $724,  answer. 

7.  The  amount  of  a  note  at  interest  for  1  yr.  4  mo.  at  6%  is 
$432.     Find  the  principal. 

Solution  :  (1)  1  yr.  4  mo.  =  i  yr. 

(2)  432=  ^(100+6x1) ^Pxl08 

100  100    * 

(3)  P=*^^  =  4O0. 
Answer,  $400. 

8.  At  what  rate  will  any  principal  double  itself  in  8  yr.  ? 

Note. — Any  principal  has  doubled  itself  when  the  interest  equals  the 
principal. 

Solution:  (1)  Interest  =  P. 

(2)P  =  ^.     (Why?) 

Answer,  121%. 

Note. —  In  (3),  the  P's  cancel. 

Another  solution  :  (1)  Int.  for  8  yr.  =  100^  of  prin.     ( Why  ?) 
i  of  (1)  =  (2)  Int.  for  1  yr.  =  12^^  of  prin.,  answer. 

9.  In  what  time  w^ill  any  principal  treble  itself  at  6%  ? 

Solution  :  (1)  Interest  =  2  P.     ( Why  ? ) 

(2)  2P=^^^—.    (Why?) 
\  J  100  ^        ^     ^ 

Answer,  88^  yr.  =33  yr.  4  mo. 


ADVANCED    ARITHMETIC.  287 

Another  solution  :  (1)  Q%  of  prin.  =  int.  for  1  yr. 
i  of  (1)  =  (2)  1%  of  prin.  =  int.  for  ^  yr. 
200 X (2)  =  (3)  200^  of  prin.  =  int.  for  33^  yr.,  answer. 

EXERCISE  CXIX. 

1.  Define  interest,  principal,  rate,  amount. 

2.  What  is  simple  interest  f 

3.  How  does  annual  interest  differ  from  simple  interest  ? 

J^.  How  does  compound  interest  differ  from  annual  interest  ? 

5.  Why  should  the  time  be  expressed  in  years  ? 

6.  Explain  the  common  interest  method  of  counting  time. 

7.  Explain  the  hankers^  interest  method. 

8.  Explain  the  exact  interest  method. 

9.  Develop  the  formulas  for  simple  interest. 

10.  Commit  to  memory  the  relations. 

Find  the  simple  interest  on  — 

11.  $240  for  2  yr.  at  6%. 

12.  $78  for  8  yr.  at  8%. 

13.  $654  for  5  yr.  at  5%. 
U.  $356  for  6  yr.  at44%. 

15.  $2400  for  4  yr.  at  8%. 

16.  $468.50  for  3  yr.  at  10%. 

17.  $763.65  for  2  yr.  at  7%. 

18.  $95000  for  12  yr.  at  5i%. 

19.  $86  for  2  mo.  at  10%. 

20.  $1500  for  3  mo.  at  8%. 

21.  $684  for  6  mo.  at  6%. 

22.  $2450.30  for  4  mo.  at  7%. 

23.  $869  for  9  mo.  at  10%. 
2J^.  $1250  for  5  mo.  at  12%. 

25.  $763.40  for  8  mo.  at  9%. 

26.  $1386.60  for  10  mo.  at  8%. 

27.  $79  for  30  da.  at  12%. 


288  ADVANCED    ARITHMETIC. 

28.  $156  for  45  da.  at  10%. 

29.  $288  for  50  da.  at  10%. 

30.  $855  for  33  da.  at  8%. 

31.  $926.42  for  15  da.  at  12%. 

32.  $1346.29  for  60  da.  at  6%. 

33.  $2400  for  93  da.  at  8%. 
3J^.   $1198.75  for  100  da.  at  7%. 

35.  $248  for  1  yr.  3  mo.  at  8%. 

36.  $192  for  2  yr.  2  mo.  at  7%. 

37.  $486  for  4  yr.  6  mo.  at  6  %. 

38.  $1573.59  for  3  yr.  4  mo.  at  8%. 

39.  $2576.42  for  5  yr.  6  mo.  at  5%. 

40.  $986.54  for  6yr.  8  mo.  at  9%. 
J^l.   $3546  for  10  yr.  9  mo.  at  4%. 
J^2.   $5684.76  for  15  yr.  8  mo.  at  3^%. 
43.   $384  for  3  mo.  15  da.  at  6%. 

U'   $576  for  2  mo.  18  da.  at  7%. 

45.  $945  for  4  mo.  27  da.  at  5%. 

46.  $128.73  for  3  mo.  20  da.  at  6%. 

47.  $894.65  for  6  mo.  12  da.  at  8%. 

48.  $1246.68  for  6  mo.  15  da.  at  8%. 

49.  $967.95  for  8  mo.  24  da.  at  7%. 

50.  $684.88  for  2  mo.  5  da.  at  10%. 

51.  $129  for  1  yr.  3  mo.  15  da.  at  5%. 

52.  $2700  for  1  yr.  4  mo.  18  da.  at  6%. 


Find  the  amount  of — 

53.  $678  for  3  yr.  3  mo.  24  da.  at  6%. 

54.  $940  for  4  yr.  3  mo.  21  da.  at  7%. 

55.  $1568  for  5  yr.  8  mo.  18  da.  at  8%. 

56.  $2568.72  for  2  yr.  9  mo.  27  da.  at  5  %. 

57.  $5468.46  for  7  yr.  10  mo.  19  da.  at  4%. 

58.  $4698.58  for  6  yr.  7  mo.  25  da.  at  5^%. 


ADVANCED    ARITHMETIC.  289 

Find  the  rate  required  for — 

59.  $800  to  produce  $114  interest  in  1  yr.  7  mo. 

60.  $540  to  produce  $64.80  interest  in  2  yr. 

61.  $276  to  produce  11.96  interest  in  6  mo.  15  da. 

62.  $1500  to  produce  $117.50  interest  in  1  yr.  6  mo.  24  da. 

63.  $450  to  produce  $12.60  interest  in  8  mo.  12  da. 

6J^.  $780  to  produce  $142.35  interest  in  1  yr.  9  mo.  27  da. 

Find  the  time  required  for — 

65.  $300  to  produce  $52.50  interest  at  5%. 

66.  $540  to  produce  $86.40  interest  at  6%. 

67.  $1000  to  produce  $110  interest  at  8%. 

68.  $240  to  produce  $16.80  interest  at  10%. 

69.  $5000  to  produce  $262.50  interest  at  7%. 

70.  $840  to  produce  $11.76  interest  at  8%. 

71.  $235  to  produce  $2,585  interest  at  12%. 

72.  $340  to  produce  $5.10  interest  at  6%. 

Find  the  principal  required  to  produce — 

73.  $30  interest  in  2  yr.  6  mo.  at  6%. 
7Jf.  $50.40  interest  in  1  yr.  9  mo.  at  8%. 

75.  $208  interest  in  4  yr.  4  mo.  at  10% . 

76.  $96.35  interest  in  1  yr.  8  mo.  15  da.  at  6%. 

77.  $35  interest  in  3  mo.  at  7%. 

78.  $28. 12 J  interest  in  4  mo.  15  da.  at  5%. 

79.  $6.40  interest  in  45  da.  at  8%. 

80.  $13.50  interest  in  90  da.  at  6%. 

Find  the  principal  lohich  amounts  to — 

81.  $570  in  2  yr.  4  mo.  at  6%. 

82.  $751.20  in  6  mo.  15  da.  at  8%. 

83.  $904.44f  in  1  yr.  3  mo.  20  da.  at  10%. 

84.  $2516.661  in  30  da.  at  8%. 


290  ADVANCED   ARITHMETIC. 

Find  the  interest  {common  interest  method)  on — 

85.  $350  from  Jan.  10,  1875,  to  June  15,  1876,  at  6%. 

86.  $500  from  May  1,  1880,  to  Dec.  1,  1884,  at  S%. 

87.  $490  from  March  15,  1897,  to  Jan.  1,  1901,  at  8%. 

88.  $875  from  July  5,  1896,  to  May  30,  1899,  at  10%. 

89.  $400  from  June  20,  1895,  to  Feb.  1,  1898,  at  6%. 

Find  the  interest  {bankers''  interest  method)  on — 

90.  $80  from  Jan.  8,  1887,  to  Feb.  20,  1887,  at  10%. 

91.  $250  from  July  15,  1890,  to  Sept.  20,  1890,  at  8%. 

92.  $385  from  Aug.  12,  1891,  to  Nov.  1,  1891,  at  12%. 

93.  $1200  from  Jan.  18,  1892,  to  April  1,  1892,  at  8%. 
9^.  $840  from  May  12,  1893,  to  Sept.  5,  1893,  at  8%. 

Find  the  exact  interest  on — 

95.  $125  from  Feb.  1,  1891,  to  Dec.  1,  1891,  at  6%. 

96.  $250  from  May  18,  1891,  to  July  31,  1891,  at  8%. 

97.  $500  from  June  15,  1893,  to  Nov.  1,  1893,  at  5%." 

98.  $475  from  Dec.  25,  1894,  to  March  19,  1895,  at  6%. 

99.  $1000  from  May  4,  1895,  to  Oct.  10,  1895,  at  7%. 

100.  $850  from  Jan.  25,  1896,  to  June  30,  1896,  at  8%. 

101.  In  what  time  will  any  principal  double  itself  at  8%? 

102.  In  what  time  will  any  principal  treble  itself  at  10%? 

103.  At  what  rate  will  any  principal  double  itself  in  12  yr.? 
lOJ^.  At  what  rate  will  any  principal  treble  itself  in  18  yr.  ? 

143.  Notes. 

[FORM.] 


No.    Ii2.  Sedan.  Kans.,  lu.l'T^    10,    i8C)S. 

OyxAAAj   '\X'VoJ\A^    after  date,     A    promise  to  pay    O.     \0. 
C^OxAaXL,    or  order,  at 

THE  FIRST  NATIONAL  BANK  OF  SEDAN. 

CJLAM^  c)=vUyTVcUuxL  CUTUcL  '^/loo  Dollars,  with  interest  at  the 
rate  of  ten  per  cent,  per  annunn  until  paid,  for  value  received. 

$500.00.  A.  d.  ']UaAX)^T\^'\^. 


ADVANCED    ARITHMETIC.  291 


A  Promissory  Note  is  a  promise  to  pay  to  the  party  designated 
a  certain  sum  of  money,  with  or  without  interest,  at  the  time 
specified  in  the  note. 

The  principal  is  the  amount  of  money  mentioned  in  the 
note.  The  Face  of  a  note  is  what  it  calls  for  on  its  face.  If  a 
note  bears  interest,  the  face  includes  principal  and  interest ;  if 
the  note  does  not  bear  interest,  the  face  is  the  principal. 

Note. — Most  authors  define  the  face  of  a  note  as  the  principal  only ; 
but,  if  that  be  so,  we  should  in  the  interest  of  simplicity  discard  the 
term  face.    ( See  Bank  Discount.) 

An  Indorsement  is  something  written  on  the  back  of  a  note. 
The  above  note  is  payable  to  E.  B.  Ferrell,  or  his  order.  He 
may  order  it  j)aid  to  W.  R.  Ferrell ;  (see  first  indorsement  above.) 
This  is  called  an  indorsement  in  full,  because  it  is  complete.  A 
holder  of  a  note  may  merely  sign  his  name  on  the  back ;  thus 
he  orders  it  paid  to  any  person  who  may  be  the  holder  at  the 
time  the  note  is  due.     This  is  called  indorsement  in  blank. 

Note.— There  are  other  forms  of  indorsing  a  note  for  transfer,  but 
they  are  not  so  commonly  used  as  the  above. 

When  payments  are  made  upon  a  note,  before  final  settle- 
ment, their  dates  and  amounts  are  placed  on  the  back  of  the 
note  (see  "Indorsements  of  payments  "  above). 


292  ADVANCED    ARITHMETIC. 

In  many  States,  the  holder  of  a  note  or  bill  due  at  a  future 
date  must  allow  8  days  more  than  the  time  stated  in  the  note 
for  payment.  These  are  called  Days  of  Grace.  The  holder  gets 
interest  for  this  extra  time,  if  the  bill  bears  interest. 

Note. — It  is  sometimes  supposed  that  days  of  grace  are  only  allowable 
in  Bank  Discount.  This  is  a  mistake.  It  is  a  law  governing  the  payment 
of  notes  and  bills,  and  not  a  principle  of  Bank  Discount.  But  remember 
that  it  applies  only  to  notes  and  bills,  and  not  to  all  debts  that  may  bear 
interest.  The  perplexing  question  to  the  student  is,  *'When  am  I  to 
include  datjs  of  grace  and  when  not  f  "  If  a  problem  reads  "  What  is  due 
Feb.  10,  1897?"  just  count  to  that  date.  If  the  problem  reads  "and 
runs  for  2  yr.  4  mo.,"  just  use  that  time.  But,  if  the  wording  of  the 
note  is  given,  you  must  allow  for  days  of  grace,  in  those  States  where 
days  of  grace  are  allowed. 

EXERCISE  CXX. 

1.  Write  a  note  that  bears  interest  from  date. 

2.  Why  should  a  note  be  dated  ? 

3.  Who  is  the  maker  of  your  note  ? 
If..  Who  is  the  payee  f 

5.  Write  a  note  that  bears  interest  after  it  is  due  only. 

6.  What  is  the  law  concerning  days  of  grace  in  your  State  ? 

7.  If  a  note  says  nothing  about  interest,  what  rate  of  inter- 
est will  it  bear  after  it  is  due,  in  your  State  ? 

8.  What  is  the  highest  rate  of  interest  you  can  legally 
charge  in  your  State  ? 

9.  If  you  charge  more  than  is  legal,  what  is  it  called  ? 

10.  What  does  the  law  say  about  usury  in  your  State  ? 

11.  What  is  indorsement  in  blank  f  in  full  f 

12.  How  are  payments  indorsed  ? 

144.  Partial  Payments.  —  When  partial  payments 
are  made  on  a  note  or  debt  bearing  interest  and  running  one 
year  or  less,  it  is  customary  to  reckon  interest  on  the  principal 
and  on  each  payment  to  the  time  of  settlement,  and  subtract 


ADVANCED    ARITHMETIC.  298 

the  amount  of  the  payments  from  the  amount  of  the  debt. 
This  is  called  ^ ^ Merchants^  Rule.''^ 

Principles  of  the  Merchants'  Rule  :  1.  The  whole  principal 
bears  interest  according  to  the  terms  of  the  note  until  final  settle- 
ment. 

2.  Each  payment  hears  interest  from  the  time  it  is  made  until 
date  of  final  settlement. 

When  the  debt  or  note  bears  interest  for  more  than  one  year, 
the  interest  is  computed  by  the  "  t/.  S.  Rule.'" 

Principles  of  the  U.  S.  Rule:  1.  Payments  must  first  dis- 
charge interest  due,  and  the  balance,  if  any,  will  be  applied  to  the 
discharge  of  the  principal. 

2.  Interest  must  not  bear  interest. 

3.  Payments  must  not  hear  interest. 

EXAMPIiES. 

1.  A  note  of  $400,  dated  Jan.  1,  1897,  interest  6%,  has  the 
following  payments:  May  1,  1897,  $30;  Nov.  16,  1897,  $100. 
What  was  due  Jan.  1,  1898  ?     (Merchants'  rule.    Why?) 

Solution :  (1)  Interest  on  $400  for  1  yr.  at  6^  =  $24. 

(2)  $400+ $24  =  $424,  amt.  of  debt  at  maturity. 

(3)  From  May  1,  1897,  to  Jan.  1,  1898  =  8  mo.  =  |  yr. 

(4)  Int.  on  $30  for  |  yr.  at  6^  =  $1.20. 

(5)  $30  +  $1.20  =  $31.20,  amt.  of  first  payment. 

(6)  From  Nov.  16, 1897,  to  Jan.  1, 1898  =  1  mo.  15  da.  =  i yr. 

(7)  Int.  on  $100  for  i  yr.  at  6^  =  $.75. 

(8)  $100+$.75  =  $100.75,  amt.  of  second  payment. 

(9)  $424 -$31.20 -$100.75  =  $292.05,  answer. 

Note. —  The  work  required  in  finding  the  interest  in  (1),  (4)  and  (7) 
need  not  appear  in  the  solution ;  but  the  teacher  should  insist  upon 
speed  and  accuracy  in  mechanical  work. 

2.  A  note  of  $150  is  dated  May  10,  1896;  interest  6%.     It 


294  ADVANCED   ARITHMETIC. 

has  the  following  indorsements:  Sept.  10,  1897,  $32;  Sept.  10, 
1898,  $6.80.     What  is  due  Nov.  10,  1898  ?    (U.  S.  Rule.    Why?) 

Dates. 

1898—11—10 

1898—  9—10  ^,  ,  ,      .  .         . 

^onj Q_io  Note. — Arrange  the  dates  in  order, 

-.ggg 5_io  viith.  the  latest  date  above.    Begin  be- 

Payments.  low ;  Subtract  each  date  from  the  one 

1_  4 0 $32  n^xt  above  it. 

1—  0—  0 $6,80 

2—  0 
Solution  :  (1)  Int.  on  $150  for  U  yr.  at  6^  =  $12. 

(2)  $150+$12-$32  =  $130,  new  principal. 

(3)  Int.  on  $130  for  1  yr.  at  6^  =  $7.80. 

(4)  $7.80 -  $6.80  =  $1,  unpaid  interest. 

(5)  Int.  on  $130  for  ^  yr.  at  6^  =  $1.30. 

(6)  $130+$1+$1. 30  =  $132.30,  due  at  maturity. 

EXERCISE  CXXI. 

1.  A  note  for  $820  was  given  June  12, 1897 ;  interest  6%.  It 
has  the  following  indorsements :  Aug.  12,  1897,  $100 ;  Nov.  12, 
1897,  $250;  Jan.  12, 1898,  $120.     What  was  due  Feb.  12,  1898  ? 

2.  A  note  of  $1200,  dated  April  1,  1890,  payable  on  demand, 
with  interest  at  7%,  has  the  following  indorsements:  May  6, 
$210;  July  5,  $210;  Oct.  18,  $822.    What  was  due  Jan.  1, 1891  ? 

3.  A  note  of  $2000,  dated  Jan.  4,  1893,  interest  6%,  has  the 
following  indorsements:  Feb.  19,  1898,  $400;  June  29,  1894, 
$1000;  Nov.  14,  1894,  $520.    What  was  due  Dec.  24,  1896  ? 

4..  Find  the  amount  due  at  maturity  on  note  No.  12  on  pages 
290-1,  including  8  days  of  grace  (Kansas  allows  days  of  grace). 

5.  A  note  of  $500,  dated  March  1,  1895,  interest  6%,  has  the 
following  indorsements :  Sept.  1,  1895,  $10;  Jan.  1,  1896,  $80; 
July  1,  1896,  $11;  Sept.  1,  1896,  $80.  What  is  due  March  1, 
1897  ? 

145.   Annual  Interest.     (  See  definition,  p.  281.) 
Example:   A  note  of  $800  drawing  interest  at  8%,  payable 


ADVANCED    ARITHMETIC.  295 

annually,  runs  for  5  years  3  months :    find  the  amount  due  at 
maturity,  if  no  payments  have  been  made. 

Note. — Each  year's  interest  will  bear  interest  after  it  becomes  due, 
as  follows : 

1st  year's  int.  for  4  yr.  3  mo. 
2d  year's  int.  for  3  yr.  3  mo. 
3d  year's  int.  for  2  yr.  3  mo. 
4th  year's  int.  for  1  yr.  3  mo. 
5th  year's  int.  for  0  yr.  3  mo. 

Total 11  yr.  3  mo. 

Therefore,  in  addition  to  the  simple  interest  on  the  principal  for  5  yr. 
3  mo.,  we  must  count  the  interest  on  1  year's  interest  for  the  sum  of  the 
above  periods,  or  11  yr.  3  mo. 

Solution:  (1)  Int.  on  $800  for  1  yr.  at  8^  =  $64,  due  annually. 

(2)  Int.  on  $800  for  5^  yr.  at  8^  =  $336,  int.  on  principal. 

(3)  Int.  on  $64  for  llj  yr.  at  8^  =  $57.60,  int.  on  int. 

(4)  $800+$336 +$57.60  =  $1193.60,  amount  due. 

EXERCISE  OXXII. 
Find  the  annual  interest  — 

1.  On  $700  for  4  yr.  3  mo.,  at  6%,  payable  annually. 

2.  On  $840  for  2  yr.  4  mo.,  at  8%,  payable  semi-annually. 

3.  On  $1000  for  6  yr.  2  mo.  15  da.,  at  10%,  payable  annually. 
4-.  On  $800  for  1  yr.  9  mo.,  at  5%,  payable  quarterly. 

5.  On  $400  for  6  yr.  4  mo.  8  da.,  at  10%,  payable  annually. 

146.   Compound  Interest.     (See  definition,  p.  281.) 

EXAMPIiES. 

1.  A  note  for  $800,  drawing  interest  at  8% ,  payable  annually, 
runs  for  5  yr.  3  mo. :  find  the  amount  due  at  maturity  by  com- 
pound interest. 

Solution:   (1)  Amt.  of  $800  at  S%  for  1  yr.  =$64 +$800  =  $864. 

(2)  Amt.  of  $864  at  S%  for  1  yr.  =  $69.12+$864  =  $933.12. 

(3)  Amt.  of  $933.12  at  8%  for  1  yr.  =  $74.6496  =  $933.12  =  $1007.77. 

(4)  Amt.  of  $1007.77  at  S%  for  1  yr.  =  $80.62 +  $1007.77  =  $1088.39. 

(5)  Amt.  of  $1088.39  at  H%  for  1  yr.  =  $87.07 +  $1088.39  =  $1175.46. 

(6)  Amt.  of  $1175.46  at  8%  for  i  yr.  =  $23.51 +$1175.46  =  $1198.97. 


296 


ADVANCED    ARITHMETIC. 


Note. — You  will  observe  that  this  is  the  same  problem  which  is  solved 
by  annual  interest.  Up  to  the  end  of  the  second  year,  the  results  ob- 
tained by  annual  interest  and  compound  interest  would  be  the  same,  but 
after  that  time  the  compound  interest  plan  begins  to  advance  faster  than 
the  annual  interest  plan.    Why  is  this  so? 

Solving  problems  as  above  would  be  too  tedious  for  use  in 
business.  A  person  having  many  problems  to  solve  in  com- 
pound interest  should  use  a  table  like  the  following : 

TABLE, 

Showing  the  amount  of  $1,  at  3,  4,  5,  6,  7  and  8  per  cent.,  at  Compound  In- 
terest, for  any  number  of  years,  from  1  to  25. 


Yr. 

3  per  cent. 

4:  per  cent. 

5  per  cent. 

6  per  cent. 

1  per  cent. 

8  per  cent. 

1 

1.03 

1.04 

1.05 

1.06 

1.07 

1.08 

2 

1.060 

1.0816 

1.1025 

1.1236 

1.1449 

1.1664 

3 

1.092727 

1.124864 

1.157625 

1 . 191016 

1.225043 

1.259712 

4 

1.125509 

1.169859 

1.215506 

1.262477 

1.310796 

1.360488 

5 

1.159274 

1.216653 

1.276282 

1.338226 

1.402551 

1.469328 

6 

1.194052 

1.265319 

1.340096 

1.418519 

1.500730 

1.586874 

7  1.229874 

1.315932 

1.407100 

1.503630 

1.605781 

1.713824 

8 

1.266770 

1.368569 

1.477455 

1.593848 

1.718186 

1.850930 

9 

1.304773 

1.423312 

1.551328 

1.689479 

1.838459 

1.999004 

10 

1.343916 

1.480244 

1.628895 

1.790848 

1.967151 

2. 158924 

11 

1.384234 

1.539454 

1.710339 

1.898299 

2.104851 

2.331638 

12 

1.425761 

1.601032 

1.795856 

2.012196 

2.252191 

2.518170 

13 

1.4685S4 

1.665074 

1.885649 

2.132928 

2.409845 

2.719623 

14 

1.512590 

1.731676 

1.979932 

2.260904 

2.578534 

2.937193 

15 

1.557967 

1.800944 

2.078928 

2.396558 

2.759031 

3.172169 

16 

1.604706 

1.872981 

2.182875 

2.540352 

2.952163 

3.425942 

17 

1.652848 

1.947900 

2.292018 

2.692773 

3.158815 

3.700018 

18 

1.702433 

2.025817 

2.406619 

2.854339 

3.379932 

3.996019 

19 

1.753506 

2.106849 

2.526950 

3.025600 

3.616527 

4.315701 

20 

1.806111 

2.191123 

2.653298 

3.207135 

3.869684 

4.660957 

21 

1.860295 

2.278768 

2.785963 

3.399564 

4.140562 

5.033833 

22 

1.916103 

2.369919 

2.925261 

3.603537 

4.430401 

5.436540 

23 

1.973587 

2.464716 

3.071524 

3.819750 

4.740529 

5.871463 

24 

2.032794 

2.563304 

3.225100 

4.048935 

5.072366 

6.341180 

25 

2.093778 

2.665836 

3.386355 

4.291871 

5. 427432 

6.848475 

2.  Find  the  compound  interest  on  $1500  for  12  yr.  at  7%, 
payable  annually. 


ADVANCED    ARITHMETIC.  297 

Solution:  (1)  Comp.  amt.  of  $1  for  12  yr.  at  7^  =  $2.252191. 
1500x(l)  =  (2)  Comp.  amt.  of  $1500  for  12  yr.  at  7^  =  $3378.2865. 
(3)  $3378.2865 -$1500  =  $1878.2865,  answer.. 

3.  Find  the  compound  amount  of  $400  for  4  yr.  2  mo.  at  6%, 
payable  semi-annually. 

Note. — Double  the  time  and  use  lot  the  rate.  Compound  interest 
for  4  yr.  2  mo.  at  Q%,  payable  semi-annually,  is  the  same  as  for  8  yr.  4  mo. 
at  3^,  payable  annually. 

Solution  :  (1)  Comp.  amt.  of  $1  at  S%  for  8  yr.  =  $1.26677. 

(2)  Comp.  amt.  of  $400  at  3%  for  8  yr.  =  $506,708. 

(3)  Int.  on  $506,708  at  S%  for  4  mo.  =  $5,067. 

(4)  $506.708+$5.068  =  $511,776,  answer. 

EXERCISE  CXXIII. 

Find  the  compound  interest  as  in  example  1 — 

1.  On  $1000  for  4  yr.  at  5%,  payable  annually. 

2.  On  $850  for  2  yr.  at  4%,  payable  semi-annually. 

3.  On  $900  for  5  yr.  4  mo.,  at  6%,  payable  annually. 

Find  the  compound  interest  as  in  examples  2  or  3 — 
J^.  On  $700  for  8  yr.  at  5%,  payable  annually. 

5.  On  $1200  for  9  yr.  at  8%,  payable  semi-annually. 

6.  On  $1100  for  6  yr.  at  12%,  payable  quarterly. 

7.  Find  the  compound  amount  on  $1200  for  5  yr.  2  mo.  at 
6%,  payable  semi-annually. 

8.  Find  the  compound  amount  on  $600  for  17  yr.  3  mo.  15 
da.  at  10%,  payable  annually. 

147.  Bank  Discount.  —  The  face  oi  a  note  is  the 
amount  due  at  maturity.  If  the  note  bears  interest,  the  face 
is  the  amount  of  principal  and  interest  due  at  maturity.  If 
the  note  does  not  bear  interest,  the  face  is  the  principal.  Bank 
Discount  is  the  simple  interest  counted  on  the  face  of  the  note 
from  the  date  of  discounting  to  the  date  of  maturity.     The 


298  ADVANCED    ARITHMETIC. 

manner  of  counting  the  interest  is  not  different  from  that  al- 
ready studied.  The  j^foceeds  are  the  difference  between  the 
bank  discount  and  the  face  of  the  note.  The  "^me  to  run''''  is 
the  actual  number  of  days  from  the  date  of  discount  to  the 
date  of  maturity;  this  is  sometimes  called  the  '■HerDi  of  dis- 
counts^'' which  is  perhaps  a  more  appropriate  name. 

To  find  the  date  of  maturity  of  a  note :  ( 1 )  When  the  time  for 
which  the  note  bears  interest  is  expressed  in  months,  as,  "due 
in  4  mo. , ' '  count  by  months.  From  a  certain  day  in  one  month 
to  the  same  day  in  the  next  month  is  1  mo.,  and  so  on.  (2) 
When  the  time  is  expressed  in  days,  count  the  actual  days  to 
find  the  date  of  maturity. 

Principles:  1.  The  face  of  the  note,  the  amount  due  at  matur- 
ity, is  the  amount  to  be  discounted. 

2.  The  hank  discount  is  the  simple  interest  on  the  face  of  the  note 
for  the  given  time  at  the  given  rate. 

3.  The  face  of  the  note  minus  the  hank  discount  equals  the  pro- 
ceeds. TERMS. 

F,  face  of  the  note. 

R,  number  of  %  of  the  discount. 

T,  time  to  run. 

D,  discount. 

P,  proceeds. 

Developing  the  formulas  for  hank  discount, 

.     r.     FRT 

{!)  P.  =  F-D.    (Prin.  3.) 

,..     r^.  r.     T.     FRT     lOOF-FRT 

il)  =  i2)  P=F-^  = 10^—'°^' 

F{100-RT) 
^'  ^-  100 

Note. — In  the  following  examples  and  exercise,  days  of  grace  are  con- 
sidered. 


ADVANCED    ARITHMETIC.  299 


examfi.es. 


1.  A  note  of  $240,  dated  Jan.  1,  1888,  time  4  months,  was 
discounted  March  5,  1888,  at  10%  :  find  the  time  to  run,  and 
proceeds. 

Solution  :  (1)  Date  of  maturity :  4  mo.  3  da.  from  Jan.  1,  or  May  4. 

(2)  Term  of  discount :  From  Mar.  5  to  May  4,  or  60  days.  =  }  yr 

(3)  pjmoo^i-io)^^ 

Proceeds,  $236. 

2.  A  note  of  $2400,  dated  July  3,  1891,  bearing  5%  interest, 
was  discounted  Jan.  9,  1892,  at  10%  ;  the  note  falls  due  10 
months  from  date.  Find  the  time  to  run,  bank  discount,  and 
proceeds. 

Solution  ;  (1)  Date  of  maturity :  10  mo.  3  da.  from  July  3,  May  6,  1892. 

(2)  Int.  on  $2400  for  Ui  yr.  at  5^  =  $101. 

(3)  $2400+$101  =  $2501,  face  of  note. 

(4)  Term  of  discount:  From  Jan.  9,  1892,  to  May  6,  1892,  or  118 

da.  (leap  year).  =  y%  yr. 
(K^  n    2501xl0x59_ 
^^^  ^-     100x180     -^^•^^- 

Discount,  $81.98. 
(6)  P=  $2400 - $81.98  =  $2419.02,  proceeds. 

3.  For  what  sum  must  a  60-day  note,  without  interest,  be 
drawn  to  produce  $500,  when  discounted  for  the  full  time  at  6%  ? 

F  {100 -TR) 


Solution:  (1)  P- 


100 
JP(100-^x6) 


(2)  500=  ^^  . 

(2)  =  (3)500  =  ^:^. 

(4)  F=^^^^^^^  =  ^.SO,. 
Face,  $505,306. 
Another  Solution  :  (1)  Bank  discount  on  any  principal  for  ^V  yr.  at  Q%  = 
laV^  of  principal. 

(2)  Proceeds:  100^  of  prin.-lgJjj^  of  prin.  =  98H^  of 

prin. 

(3)  98Jg^of  prin.=$500. 

(4)  1^  of  prin.  =  $5.05306. 

(5)  100^  of  prin.  =$505,306,  answer. 


300  ADVANCED    ARITHMETIC. 

EXERCISE  CXXIV. 

1.  Commit  to  memory  the  principles  of  bank  discount. 

2.  Develop  the  formulas,  and  give  the  relations. 

Find  date  of  maturity,  time  to  run,  and  hank  discount. 

3.  Prin.,  $600;  date,  May  80;  time,  5  mo.;  rate,  10%  ;  dis- 
counted, July  1. 

4.  Prin.,  $840;  date,  Feb.  15;  time,  8  mo.;  rate,  0%;  dis- 
counted, June  10. 

5.  Prin.,  $1000;  date,  July  10;  time,  90  da.;  rate,  8%  ;  dis- 
counted, Aug.  1. 

Find  date  of  maturity,  time  to  run,  and  jiroceeds,  if  each  of  the 
following  hears  8%  interest  from  date  : 

6.  Prin.,  $800;  date,  Jan.  10;  time,  6  mo. ;  rate,  10%  ;  dis- 
counted, Mar.  6. 

7.  Prin.,  $500;  date,  Apr.  12;  time,  9  mo.;  rate,  6%;  dis- 
counted, Aug.  18. 

8.  Prin.,  $470;  date,  May  15;  time,  120  da.;  rate,  10%; 
discounted,  July  1. 

9.  I  desire  to  use  to-day  $080,  which  I  can  secure  by  giving  a 
bankable  note  payable  in  80  days,  without  interest,  discounted 
at  9%.     For  what  sum  must  I  write  the  note  ? 

148.  True  Discount. — The  Present  Worth  of  a  debt 
payable  at  a  future  date  without  interest  is  that  sum  of  money 
which,  when  put  at  interest  for  the  given  time,  will  amount  to 
the  debt.  The  difference  between  the  present  worth  and  the 
debt  is  the  True  Discount.  To  find  the  present  worth  of  a  debt 
is  the  same  problem  that  we  had  in  Interest :  to  find  the  prin- 
cipal when  the  amount  of  the  principal  and  the  interest  are 
given.     (See  formula  2,  p.  284.) 

Note. — If  the  debt  discounted  be  a  note,  given  in  a  State  where  days 
of  grace  are  recognized  by  law,  days  of  grace  should  be  considered  in  true 
discount  as  well  as  in  any  other  application  of  interest. 


ADVANCED   ARITHMETIC.  301 

If  a  debt  or  note,  bearing  interest,  is  discounted,  the  interest 
due  at  maturity  must  be  added  to  the  principal,  before  dis- 
counting. 

EXAMPIiES. 

1.  What  is  the  present  worth  and  true  discount  of  a  debt  of 
$627  if  paid  9  months  before  it  is  due,  discount  6%? 

P{im+RT) 


Solution :  (1)  A  = 
(2)  627= 


100 
P{100+6Xj%)  ^PxlO^i 
100  100 


(3)  P=?27xlOOx_2^gQo^ 

209 

(4)  Present  Worth  =  ^Q0O. 

(5)  True  Discount  =  ^27. 


No.  5.  Topeka,   Kans.,  KX/TU.  5,    '900. 

Ui/V-V    TnxvTbtnA/     after    date,     J     promise    to    pay 

CaxiuTlV     ^     Co., or  order, 

c3umX-\K^  jiuyTLcLAyexl^  cutixL    "-^/wo  Dollars, 

with  interest  from  date  at  the  rate  of    5    per  cent,  per  annum,  for 
value  received. 

$  1200.00.  .  c^.  Q..  (XclcuTn^. 


2.  Find  the  true  present  worth  of  note  No.  5,  on  April  9,  1900, 
discount,  6%. 

Solution  :  (1)  Interest  on  $1200  at  5%  for  5  mo.  3  da.  =$25,50. 

(2)  $1200 +$25. 50  =$1225.50. 

(3)  5  mo.  3  da.  from  Jan.  5,  1900,  is  June  8,  1900,  the 

date  of  maturity. 

(4)  From  April  9  to  June  8  =  60  da.,  the  term  of  dis- 

count. 

(5)  1225.50  =  -P^^Q^+^><^). 

(6)  p^jjg5.5x  100 ^-^213.366. 
Answer,  $1213.366. 


302  ADVANCED    ARITHMETIC. 

EXERCISE  OXXV. 

1.  A  debt  of  $942  is  due  in  4yr.  6  mo.  6  da.,  without  interest : 
find  its  present  worth  if  money  is  worth  6%. 

2.  A  debt  of  $273.75  is  due  in  1  yr.  7  mo.,  without  interest: 
find  its  present  worth  and  true  discount  at  6%. 

3.  A  debt  of  $866  is  due  in  8  yr.  8  mo.,  without  interest :  find 
its  true  discount  at  6%. 

^.  A  debt  of  $888.20  is  due  in  5  yr.  7  mo.  27  da.,  without  in- 
terest.    Wliat  is  its  present  worth,  money  being  worth  10%? 

5.  A  note  of  $800,  dated  Jan.  10,  time,  6  mo.,  bearing  interest 
at  8%,  is  discounted  at  true  discount  Mar.  6,  at  10%.  Find 
the  true  discount. 

6.  The  difference  between  true  and  bank  discount  is  equal  to  the 
simple  interest  on  the  true  discount  for  the  given  time  at  the  given 
rate.  Illustrate  this  fact  by  solving  and  comparing  result  in 
the  following: 

What  is  the  difference  between  the  true  and  bank  discount 
on  $510  for  4  mo.  at  6%? 

149.   Exchange. — Exchange  is  the  process  of  mak- 
ing payment  in  distant  places  without  transferring  the  money. 
Exchange  is  effected  by  means  of  drafts  or  bills  of  exchange. 

[A  DRAFT.] 


No.  C|.  FIRST  NATIONAL  BANK  OF  SEDAN. 

Sedan,  Kansas,  wXl^  b,    '8Cj5. 
Pay  to  the  order  of   CJruxhlAJi^  QLoAjfO,  $i250.00 

c5iAM>  c^UyTvdyuxl  3"Lf/ta^  arnxl ]^   Dollars. 

To  WESTERN  NATIONAL  BANK,  ^.    c3.     DVC»xLu/l^, 

New  York,  N.  Y.  Cashier. 


This  is  the  form  of  an  ordinary  draft,  or  bill  of  exchange, 


ADVANCED   ARITHMETIC.  303 

issued  by  one  bank  upon  another.  This  draft  purports  to  have 
been  issued  by  the  cashier  of  the  First  National  Bank  of  Sedan 
upon  the  Western  National  Bank  of  New  York.  The  New  York 
bank  is  hereby  ordered  to  pay  $250  to  Charles  Clark. 

To  illustrate  the  use  of  a  draft:  Suppose  that  you  owe 
Charles  Clark  $250,  and  that  he  lives  near  New  York  city. 
One  of  the  easiest  as  well  as  the  safest  ways  to  pay  the  debt  is 
to  go  to  your  bank  and  get  a  draft  on  New  York,  just  like  the 
above,  and  send  it  to  Clark.  He  can  take  it  to  the  Western 
National  Bank,  or  any  other  bank  in  that  country,  and  get  his 
money. 

Banks  usually  charge  a  premium  for  issuing  drafts,  as,  i%, 
i%,  1%,  li%.  Sometimes  a  bank  will  issue  drafts  at  a  dis- 
count; that  is,  give  you  something  in  order  to  get  you  to  take 
the  drafts.  This  is  when  there  is  a  money  panic  at  the  money 
centers,  or  a  scarcity  of  money  at  the  local  bank. 

The  above  is  a  sight  draft;  it  is  payable  on  demand.  Occa- 
sionally a  customer  wants  a  draft  and  does  not  care  to  have  it 
paid  for  30,  60,  or  90  days.  Such  a  draft  is  called  a  time  draft, 
and  the  purchaser  gets  interest  on  his  money  for  that  time. 

Domestic,  or  Inland  Exchange  is  exchange  between  people  in 
the  same  nation  or  country.  Foreign  exchange  is  exchange  be- 
tween people  in  different  nations  or  countries. 

Principles:  1.  The  premium  or  discount  is  some  number  of 
per  cent  of  the  face  value  of  the  draft. 

2.  In  sight  drafts,  the  cost  is  equal  to  the  face  of  the  draft  plus 
the  premium  or  minus  the  discount. 

3.  In  time  drafts,  the  cost  is  equal  to  the  face  of  the  draft  minus 
the  interest,  plus  the  premium  or  minus  the  discount. 

terms. 
F,  the  face  of  the  draft. 
p,  number  of  %  of  premium  or  discount. 
P,  the  premium. 


304"  ADVANCED   ARITHMETIC. 

Z),  the  discount. 

(7,  the  cost. 

/,   the  interest. 

jK,  number  of  %  of  the  interest. 

y,  the  time. 

Development  of  formulas  for  exchange. 


2.    D: 


100 
pF 


100 
In  sight  drafts  at  a  premium, 

100  100       '       ' 

3     ^^F{m-{-p) 

100 
In  sight  drafts  at  a  discount, 

4    ^^F(lOO-p) 

100 
In  time  drafts  at  a  premium, 

(1)  C=F-I+P=F-™^^^. 

^  100      100  - 

g    ^^J^(100-i?r+j^) 

100 
In  time  drafts  at  a  discount, 

^       i:t      r      «      rr     FET     pF 

100 
Note. —  Days  of  grace  are  usually  considered  in  time  drafts. 


ADVANCED    ARITHMETIC. 
EXAMPIiES. 

1.  What  will  a  draft  on  Kansas  City  for  $880  cost,  at  i% 

premium?  ssOxlOOj:    ,,.,  , 

Solution:  C  = ^  =  882.2. 

100 
Answer,  $882.20. 

2.  What  will  a  draft  on  St.  Louis  for  $1200  cost,  at  i%  dis- 
count? 1200x99^    ,,_. 

Solution  :  C  = =  1194. 

100 
Answer,  $1194. 

3.  I  have  $2000 :  how  large  a  draft  can  I  purchase  with  it,  if 
I  have  to  pay  i%  premium  ? 

Solution:  (1)  2000=^^ ^^^ 
100 

(2)  F=?0«0^  =  1990.05. 
lOOi 
Answer,  $1990.05. 

4..  How  large  a  draft  can  I  buy  for  $2000,  discount  i%? 

Solution:  (1)  2000  =  ^^^^^ 
100 

(2)i.=  ?2^^  =  2005.01  +  . 
99  4 

Answer,  $2005.01+. 

5.  I  want  a  60-day  draft  on  New  York  for  $2000 :  what  will 
it  cost  me,  if  interest  is  6%  and  premium  on  New  York  is  i%? 

Solution:  a)C  =  ^^^^^^^^^±^. 
100 
2000(100-^x6+1) 

^^^^"  100  • 

2000  X  99^^^2000  X  1989^ 

'  100  100x20 

Answer,  $1989. 
Another  solution  :  (!)  Int.  on  draft  for  1  yr.  at  6^  =  6j^of  draft. 

^  of  (1)  =  (2)  Int.  on  draft  for  -i^  yr.  at  6^  =  l^V^  of  draft. 

(3)  100^  of  draft +i^  of  draft -l^V^  of  draft  =  99^o3f 

of  draft. 

(4)  100^  of  draft  =  $2000.     (Basis.) 
jU  of  (4)  =  (5)  1%  of  draft  =  $20. 

99^  X  (5)  =  (6)  99^x5%  of  draft  =  $1989,  answer. 


306  ADVANCED    ARITHMETIC. 

6.  What  will  the  draft  in  Example  5  cost  if  the  exchange  on 

9 

Note. — This  is  the  same  as  above,  except  that  you  should  use  formula 
No.  6  instead  of  No.  5.    Answer,  $1969. 

7.  How  large  a  30-day  draft  can  I  get  for  $2000,  interest  be- 
ing 8%  and  premium  i%? 

Solution:  m  2m  =^^^^^^i^±^. 

(1)    ("12000     ^^99»--^x59n 
(1)-W2000-      ^^     -100x60 

(3)  i.=  ?000^M^  =  2009.71+. 
0971 

Answer,  $2009.71  +  . 

8.  What  will  be  the  size  of  the  draft  in  Example  7,  if  ex- 
change is  at  a  discount  of  i%  ? 

Note.— This  is  the  same  as  above,  except  that  you  should  use  formula 
No.  6  instead  of  No.  5.     Answer,  $1980.52+. 

9.  What  will  an  exchange  on   London   for  £240   cost,  at 

$4.85? 

Solution:  (1)  Cost  of  £1  =  $4.85.     (Basis.) 
240x(l)  =  (2)  Cost  of  £240  =  $1164. 

10.  I  have  $2635.20  with  which  to  buy  a  draft  on  Liverpool, 
at  $4.88 :  what  will  be  the  size  of  the  draft  ? 

Solution:  (1)  $4.88  =  cost  of  £1.     (Basis.) 
^-^of(l)  =  (2)$l  =  costof£^g. 
2635.2  X  (2)  =  (3)  $2635.20  =  cost  of  £540. 

EXERCISE  CXXVI. 

1.  Define  exchange,  domestic  exchange,  foreign  exchange. 

2.  Draw  a  draft,  using  amount,  names,  date,  etc.,  as  you  may 
choose. 

3.  Develop  all  the  formulas  for  exchange. 


I 


ADVANCED    ARITHMETIC.  807 

Jf..  Commit  to  memory  the  principles. 

5.  Give  the  relations  expressed  by  the  formulas. 

6.  What  will  be  the  cost  of  a  draft  on  Chicago  for  $8600  at 
i%  premium  ? 

7.  What  will  be  the  cost  of  a  draft  on  New  Orleans  for  $9400 
at  i%  discount? 

8.  I  paid  $5420  for  a  draft  on  Kansas  City  at  |%  premium : 
what  was  the  size  of  the  draft  ? 

9.  I  paid  $7240  for  a  draft  on  New  York  at  i%  discount: 
what  was  the  face  of  the  draft  ? 

10.  What  will  be  the  cost  of  a  60-day  draft  on  San  Francisco 
for  $6540,  premium  i%,  interest  8%. 

11.  What  will  be  the  cost  of  a  90-day  draft  on  Philadelphia 
for  $2450,  discount  ^%,  and  interest  6%? 

12.  I  have  $7800  with  which  I  desire  to  purchase  a  60-day 
draft  on  New  York:  if  exchange  is  at  1^%  premium,  and 
money  is  worth  8%,  what  will  be  the  face  of  my  draft  ? 

13.  What  size  bill  of  exchange  can  I  get  on  London  for 
$12150,  if  exchange  is  $4.86? 

14-.  I  bought  a  bill  of  exchange  on  Liverpool  for  £842 :  what 
did  it  cost  if  exchange  is  $4.83^  ? 

15.  A  draft  on  Pittsburg  cost  $1771,  exchange  |%  premium: 
what  was  the  face  of  the  draft  ? 

16.  The  face  of  a  80-day  draft  on  St.  Paul  is  $4220:  what 
did  it  cost  if  exchange  is  1^%  premium  and  interest  8%  ? 

17.  A  grain  merchant  in  Toledo  sold  11875  bu.  of  corn  at  40/ 
per  bushel;  deducting  8%  commission,  he  purchased  a  60-day 
draft  with  the  proceeds,  interest  6%,  premium  2%:  required 
the  face  of  the  draft. 

18.  An  agent  owed  his  principal  $1011.84.  He  bought  a 
draft  with  the  sum,  and  remitted :  the  principal  received  $992. 
Find  the  rate  of  exchange. 

19.  What  is  the  cost  of  a  bill  on  Amsterdam  for  4800  guild- 
ers, quoted  at  41  J/,  brokerage  i  ? 


308  ADVANCED    ARITHMETIC. 

20.  Jones,  of  St.  Louis,  has  a  debt  of  $6000  to  pay  in  New- 
York.  The  direct  exchange  is  -1%  premium;  but  exchange  on 
Philadelphia  is  i%  premium,  and  from  Philadelphia  to  New 
York  is  ^%  discount.  By  circular  exchange  how  much  will 
pay  the  debt,  and  how  much  does  he  gain  over  the  direct  ? 

C.     MECHANICS. 

150.  Force. — Force  is  that  which  tends  to  produce  or 
overcome  motion  of  matter. 

The  several  units  of  force  in  common  use  are  the  poundal^ 
dyne,  pound,  and  gram,  and  may  be  defined  as  follows : 

1.  The  force  required  to  add  1  ft.  in  1  sec.  to  the  velocity  of  1  lb. 
of  matter  =  1  poundal  (pi.). 

2.  The  force  required  to  add  1  cm.  in  1  sec.  to  the  velocity  of  1  g. 
of  matter  =  1  dyne. 

3.  The  force  required  to  add  32.16  ft.  in  1  sec.  to  the  velocity  of 
1  lb.  of  matter =1  pound  (N.  Y.). 

4-.  The  force  required  to  add  980  cm.  in  1  sec.  to  the  velocity  of  1 
g.  of  matter  — 1  gram  (N.  Y.). 

Note. — It  should  be  remembered  that  we  have  pounds  of  matter  and 
pounds  of  force,  and  grams  of  matter  and  grams  of  force.  The  pupil 
should  distinguish  clearly  between  the  two. 

The  units  which  have  "N.  Y."  after  them  are  called  gravity  units,  be- 
cause they  vary  with  the  attraction  of  gravitation  at  different  places  on 
the  earth.  As  given  above,  the  units  are  correct  at  the  sea-level  at  New 
York.  The  dyne  and  poundal  are  called  absolute  units,  and  are  the  same 
for  all  places. 

TABLE    OF   EQTJrVAIiENTS. 

1  lb.  offeree  (N.  Y.)  =  32.16  pi. 

1  g.  of  force  (N.  Y.)=980  dynes. 
1  pl.  =  13826  dynes. 

EXERCISE  CXXVII. 

1.  Reduce  584  pi.  to  dynes. 

2.  Reduce  840  lb.  (N.  Y.)  to  pi. 

3.  Reduce  34  g.  (N.  Y.)  to  dynes. 


ADVANCED    ARITHMETIC.  809 

4.  Reduce  103675  dynes  to  pi. 

5.  Reduce  19296  pi.  to  lb.  (N.  Y.). 

6.  Reduce  8859200  dynes  to  lb.  (N.  Y.). 

7.  Reduce  1960  lb.  (N.  Y.)  to  g.  (N.  Y.). 

8.  Reduce  84  lb.  (N.  Y.)  to  dynes. 

The  amount  of  velocity  added  per  second  to  the  motion  of  a 
body  is  called  its  Acceleration. 

TERMS. 

F,  the  force. 

ilf,  the  mass. 

-y,  the  velocity, 

t^  the  time  (in  sec). 

a,  the  acceleration. 

Development  of  the  formulas  for  force. 

If  a  body  starts  to  move  from  a  state  of  rest,  its  velocity  at  the  end  of 
the  Ist  sec.  will  be  the  same  as  its  acceleration ;  or 
(1)  Velocity  for  1  sec.  =  a. 
and  (2)  Velocity  for  i  sec.  =  a< ;  or, 

1.  v=at. 

From  the  definition  of  a  poundal, 

(1)  Force  required  to  add  1  ft.  in  1  sec.  to  the  velocity  of  1  lb.  of  mat- 
ter =1  pi. 

(2)  Force  required  to  add  v  ft.  in  1  sec.  to  the  velocity  of  1  lb.  of  mat- 
ter-v  pi. 

(3)  Force  required  to  add  v  ft.  in  1  sec.  to  the  velocity  of  if  lb.  of  mat- 
ter =3/y  pi. 

(4)  Force  required  to  add  v  ft.  in  t  sec.  to  the  velocity  of  M  lb.  of  mat- 
ter =  —  pi. ;  or,  stated  abstractly, 

2.  F=^. 

z 

From  formula  1, 

(1)  a  =  f 

V 

Substituting  a  for      in  formula  2,  we  have  — 

3.  F=Ma. 


310  ADVANCED    ARITHMETIC. 

When  if  is  expressed  in  lb.,  v  or  a  in  ft.,  and  t  in  seconds,  formulas  2 
and  3  give  the  force  in  poundals.  When  M  is  expressed  in  g.,  v  or  a  in 
cm.,  and  t  in  sec,  these  formulas  give  the  force  in  dynes. 

Since  a  poundal  is  g^^g  POund  of  force  (N,  Y.),  and  a  dyne  is  -g^gram  of 
force  (N.Y,),  these  f ormulas  will  represent  22^^  as  m&nj  pounds  as  pound- 
als; or  Yh  as  many  grams  as  dynes. 

Now,  32.16  ft.  =980  cm. 

If  we  represent  this  amount  by  g,  these  formulas,  when  used  to  find 
the  force  in  pounds  or  grams,  will  be  — 

4.  F= — ,  and 

5.  F=^. 

9 

Note. — Do  not  forget  that  g  in  feet  is  32.16;  in  centimeters  it  is  980. 

EXAMFIiES. 

1.  What  force  will  give  15  lb.  of  matter  an  acceleration  of 
25  ft.  per  second  ? 

Solution :  F=lbx2b  =  375. 

.'.  the  required  force  is  375  poundals. 

2.  What  force  will  give  to  21  grams  of  matter  an  accelera- 
tion of  10  centimeters  per  second  ? 

Solution:  1^=21x10  =  210. 

.*.  the  required  force  is  210  dynes. 

3.  What  force  will  give  6  cwt.  20  lb.  of  matter  an  accelera- 
tion of  1  ft.  6  inches  per  second  ? 

Solution :  (1)  6  cwt.  20  lb.  =  620  lb. 

(2)  1ft.  6in.  =  lHt. 

(3)  jP=620x  11  =  930. 

.*.  the  required  force  =  930  poundals. 

4-.  What  is  the  force  in  poundals  that  will  in  10  minutes 
produce  a  velocity  of  a  mile  a  minute  in  100  lb.? 


ADVANCED   ARITHMETIC.  311 

Solution:  (1)  1  mi.  =5280  ft. 

(2)  1  min.  =  60  sec. 

(3)  10  min.  =  600  sec. 

(4)  Velocity  of  1  mi.  per  min.  =  velocity  of 
88  ft.  per  sec. 

100X88^ 
'  600  ^ 

.'.  the  reqd.  force  is  14|  poundals. 

6.  A  force  of  30  dynes,  acting  for  12  seconds  upon  a  body, 
gives  it  a  velocity  of  120  cm.  per  second.  Find  the  mass  of  the 
body. 

Solution  :  (1)  30  =  i?^^=  10  x  3f . 
12 

(2)  3f=fg  =  3. 

.*.  the  required  mass  is  3  grams. 

6.  A  force  of  40  poundals  is  acting  upon  a  mass  of  16  pounds ; 
how  much  velocity  is  given  the  body  per  second  ? 

Note. —  Acceleration  is  wanted. 

Solution:  (1)  40  =  16xa. 
(2)a  =  f^  =  2i. 
.'.  the  reqd.  acceleration  is  2|  ft.  per  sec. 

7.  A  body,  acted  upon  by  a  force  of  100  dynes,  receives  an 
acceleration  of  20  cm.  per  second.     Find  its  mass. 

Solution:  (1)  100  =  20x.3f. 
(2)  3/= -1^0- =  5. 
.'.  the  reqd.  mass  is  5  grams. 

8.  A  force  of  360  poundals  acts  upon  a  mass  of  6  pounds  for 
4  seconds.     What  velocity  did  the  body  receive  ? 

o    7    ^-  /1\    nor,      6XV      3X1' 

Solution  :  (1)  360  = = . 

(2)  ,  =  §60x2^240. 
3 

.".  the  reqd.  velocity  is  240  ft.  per  sec.  • 


312  ADVANCED   ARITHMETIC. 

9.  A  body  of  30  pounds  received  a  velocity  of  480  ft.  per  sec- 
ond from  a  force  of  120  poundals.  How  long  did  the  force  act 
to  do  this  ? 

Solution:  (1)  120  =  ^^^^^^. 
t 

(2)  120x<  =  480x30. 

.*.  the  reqd.  time  is  120 sec,  or  2  min. 

10.  How  many  pounds  of  force  will  add  to  821.6  pounds  of 
matter  a  velocity  of  20  ft.  per  second  ? 

o  ,  ,.         ^    321.6x20    ,^^ 

Solution:  F= —  =  200, 

32.16 

.*.  the  reqd.  force  is  200  lb. 

11.  A  force  of  800  pounds  gives  a  body  a  velocity  of  80  ft. 
per  second  in  5  seconds.     Find  its  weight. 

80x3f 


Solution:  (1)  800  = 


32.16x5 


3^^800X32.16X5^^^^3^ 
80 
/.  the  reqd.  weight  is  1608  lb. 

12.  Find  the  acceleration  per  second  given  to  a  body  of  700 
grams  by  a  force  of  140  grams. 

Solution:  (1)  140  =  ^^-. 
980 

140x980_ 

^^^^-~76o~-^^^- 

.'.  the  reqd.  acceleration  is  196  cm.  per  sec. 

EXERCISE  CXXVIII. 

1.  What  is  force  ?    Velocity?    Acceleration? 

2.  What  is  a  poundal  ? 

3.  What  is  a  dyne  ? 

4.  What  is  a  lb.  of  force  (N.  Y.)? 

5.  What  is  a  gram  of  force  (N.  Y.)? 


ADVANCED   ARITHMETIC.  818 

6.  Repeat  the  table  of  equivalents. 

7.  How  would  you  reduce  grams  to  pL? 

8.  How  would  you  reduce  pounds  to  dynes  ? 

9.  What  is  the  difference  between  a  pound  of  force  and  a 
pound  of  matter  ? 

W.  Develop  the  formulas  for  force. 

11.  Give  the  relation  expressed  by  each  formula. 

12.  A  constant  force  acting  upon  a  mass  of  15  g.  for  4  sec. 
gives  it  a  velocity  of  20  cm.  per  sec.     Find  the  force  in  dynes. 

13.  What  is  the  force  in  poundals  that  will  give  to  1000  lb. 
of  matter  in  20  min.  a  velocity  of  2  miles  a  minute  ? 

Note.— Express  the  time  in  sec.  and  the  velocity  in  ft.  What  is  the 
velocity  per  sec,  if  the  mass  is  traveling  at  the  rate  of  a  mile  a  minute  ? 

14'  A  mass  of  15  lb.  lying  on  a  smooth,  flat  table  is  acted 
upon  by  a  force  of  60  pi.  Find  the  velocity  at  the  end  of  2 
sec. 

15.  A  force  of  1000  dynes  acting  on  a  mass  for  1  sec.  gives  a 
velocity  of  20  cm.  per  sec.     Find  the  mass  in  grams. 

16.  What  force  in  dynes  will  in  1  sec.  give  to  a  mass  of  12  g. 
a  velocity  of  6  cm.  per  sec? 

17.  What  velocity  will  be  produced  by  a  force  of  490  dynes 
acting  on  a  mass  of  70  g.  for  10  sec? 

18.  In  what  time  will  a  force  of  7500  pi.  give  to  a  mass  of 
800  lb.  a  velocity  of  250  ft.  per  sec? 

151.  Work. — Work  is  moving  a  body ;  and  the  amount 
of  work  depends  upon  the  amount  of  force  (resistance)  over- 
come and  the  distance  through  which  that  force  is  overcome. 

The  several  units  of  work  in  common  use  are  the  foot-poundal 
(ft. -pi.),  erg ,  foot-pound  (ft. -lb.),  and  gram-centimeter  (g. -cm.), 
and  are  defined  as  follows : 

1.  The  ivork  done  in  overcoming  1  poundal  of  force  through  a 
distance  of  1  foot  —  1  foot-poundal. 


314  ADVANCED   ARITHMETIC. 

2,  The  work  done  in  overcoming  1  dyne  of  force  through  a  dis- 
tance of  1  centimeters^  1  erg. 

3.  The  work  done  in  overcoming  1  pound  of  force  through  a  dis- 
tance of  1  foot  =  l  foot-pound. 

Jf.  The  work  done  in  overcoming  1  gram  of  force  through  a  dis- 
tance of  1  centimeter  ^=1  gram-centimeter. 

Note. — The  foot-poundal  and  the  erg  are  called  absolute  units  of  work ; 
and  the  foot-pound  and  gram-centimeter  are  called  gravity  units.  This 
classification  agrees  with  that  of  the  force  units  (see  p.  308). 

TABIiE  or  EQITIVAIiENTS. 


1  g.-cm.  (]Sr.  Y.)  =  980  erg:s. 
1  ft.-lb.  (N.  Y.)  =  32.16  ft.-pl 
1  ft.-pl.=421402  ergs. 


EXERCISE  CXXIX. 

1.  Reduce  16  ft.-pl.  to  ergs. 

2.  Reduce  182  ft.-lb.  to  ft.-pl. 

3.  Reduce  842  g.-cm.  to  ergs. 

4.  Reduce  8428040  g.-cm.  to  ft.-pl. 
6.  Reduce  64820  ft.-pl.  to  ft.-lb. 

6.  Reduce  69580  ergs,  to  g.-cm. 

7.  Reduce  742  ft.-lb.  to  grgs. 

8.  Reduce  21070100  ergs,  to  ft.-pl. 

TERMS. 

ir,  the  work. 

F,  the  force  (resistance), 

d,  distance. 

Developing  the  formula  for  ivork. 
From  the  definition  of  foot-poundal, 


ADVANCED    ARITHMETIC.  315 

(1)  Work  done  in  overcoming  1  pi.  of  force  through  a  distance  of  1  ft. 
=  lft.-pl. 

(2)  Work  done  in  overcoming  F  pi.  of  force  through  a  distance  of  1  ft. 
=  i^ft.-pl. 

(3)  Work  done  in  overcoming  F  pi.  of  force  through  a  distance  of  d  ft. 
=  Fd  ft.-pl.  ;  or,  stated  abstractly, 

W=Fd. 

Note. — When  F  is  poundals  and  cZ,  ft.,  the  work  18  foot-poundals ;  when 
F  is  dynes  and  d,  cm.,  the  work  is  ergs;  when  F  is  pounds  and  d,  ft.,  the 
work  is  foot-pounds ;  and  when  F  is  grams  and  d,  cm.,  the  work  is  gram- 
centimeters. 

EXERCISE  CXXX. 

1.  Define  work,  foot-poundal,  foot-pound,  erg,  and  gram- 
centimeter. 

2.  Does  the  question  of  time  enter  into  the  work  formula  ? 
Is  the  work  in  No.  4  the  same  whether  done  in  10  min.  or  in  10 
hr.? 

3.  Develop  the  work  formula,  and  give  the  relation  which  it 
expresses. 

4-.  A  team  of  horses  draws  a  wagon  2  miles.  If  the  draft  is 
1200  pL,  how  much  work  is  performed  ? 

5.  240000  ergs  of  work  is  performed  in  overcoming  a  force  of 
800  dynes.     Through  what  distance  did  the  body  move  ? 

6.  9640  ft.-pl.  of  work  overcomes  a  certain  force  through  a 
distance  of  120  ft.  6  in.     Find  the  force. 

7.  Two  horses  draw  a  machine  40  rd.,  and  thus  do  105600 
ft. -lb.  of  work.  What  is  the  draft  of  the  machine  in  pounds  ? 
in  poundals? 

8.  How  much  work  is  done  in  lifting  50  Kg.  of  matter  20 
meters  high  ?     (Answer  in  g.-cm.) 

9.  6240  g.-cm  of  work  is  performed  in  overcoming  a  resist- 
ance of  120  dynes.     Find  the  distance. 

152.   Activity  or  Po^ver. — The  rate  at  which  work  is 


316  ADVANCED   ARITHMETIC. 

done  is  called  Power  or  Activity,  and  its  amount  depends  upon 
the  work  done  and  the  time  in  which  it  is  done. 

There  are  two  units  of  activity  in  use,  the  watt  and  the 
horse-power,  which  may  be  defined  aS  follows : 

1.  The  activity  required  to  do  10000000  ergs  of  ivork  in  1  second 
=  1  ivatt. 

2.  The  activity  required  to  do  550  ft. -lb.  of  ivork  in  1  second  =  \ 
horse-power  (h.-p.) 

Note. — The  watt  is  called  the  absolute  unit  of  activity,  and  the  horse- 
power the  gravity  unit. 

EQUIVAIiENTS. 

1  }i.-p.  (N.  Y.)=T46  watts. 

TERMS. 

A,  the  activity  or  power. 
TT,  the  work. 
t,  the  time. 

Development  of  formulas  for  activity. 

From  the  definition  of  watt, 

(1)  Activity  reqd.  to  do  10000000  ergs  of  work  in  1  sec.  =  1  watt. 

(2)  Activity  reqd.  to  do  1  erg  of  work  in  1  sec.  =  j^qqqqqqq  ^^^t. 

(3)  Activity  reqd.  to  do  W  ergs  of  work  in  1  sec.  =  j^qqqqqqq  watts. 

W 

(4)  Activity  reqd.  to  do  W  ergs  of  work  in  t  ^^<^-  =  ;^qqqqqqq^  ^2itt^ ;    or, 

stated  abstractly. 

This  formula  is  correct  for  finding  the  activity  in  watts.  By  using  the 
definition  for  horse-power,  the  formula  for  finding  horse-powers  will  be 
found  to  be  — 

^-  ^~550xC 


ADVANCED    ARITHMETIC.  317 

EXERCISE  OXXXI. 

1.  What  is  power  or  activity  f    How  does  it  differ  from  workf 

2.  Define  watt^  and  horse-power. 

3.  Reduce  520  h-p.  to  watts. 
^.  Reduce  17158  watts  to  h-p. 

5.  Develop  all  the  formulas  for  activity.  Give  the  relation 
expressed  by  each. 

6.  How  many  watts  of  activity  will  do  720000000  ergs  of  work 
in  6  sec? 

7.  How  much  work  will  be  done  by  an  activity  of  12  h-p.  in 
I  min.? 

8.  How  much  time  will  be  required  for  an  activity  of  5  watts 
to  do  150000000  ergs  of  work  ? 

9.  How  many  h-p.  of  activity  will  do  385000  ft.-lb.  of  work 
in  5  sec? 

10.  What  is  the  horse-power  of  an  engine  that  can  raise  2376 
lb.  1000  ft.  in  3  min.? 

11.  How  far  can  a  2-horse-power  engine  raise  5  tons  in  ^0 
sec? 

12.  How  long  will  it  take  a  2-horse-power  engine  to  raise  5 
tons  100  ft.? 

153.  Simple  Macliines. —  The  simple  machines  are 
the  Lever,  Wheel  and  Axle,  Inclined  Plane,  and  Screw.  (See  Dic- 
tionary or  Physics  for  definitions.) 

I.  The  elements  involved  in  lever  problems  are  power  (P), 
weight  (W),  the  distance  from  the  fulcrum  to  the  power  (Z>), 
and  the  distance  from  the  fulcrum  to  the  weight  (d). 

Note. — The  distances  D  and  d,  are  often  spoken  of  as  power  arm  and 
weight  arm,  respectively. 

Law  of  the  Lever  :  The  power  is  to  the  weight  as  the  weight 
arm  is  to  the  power  arm. 

Proportion  :  P :  W ::  d  :  D. 


318  ADVANCED    ARITHMETIC. 

II.  The  elements  involved  in  the  problems  of  the  ivheel  and 
axle  are  power  (P),  weight  (W),  radius  (diameter  or  circum- 
ference) of  the  wheel  (R),  and  the  radius  (diameter  or  cir- 
cumference) of  the  axle  (r). 

Law  of  the  Wheel  and  Axle  :  The  power  is  to  the  weight  as 
the  radius  {diameter  or  circumference)  of  the  axle  is  to  the  radius 
{diameter  or  circumference)  of  the  wheel. 

Note. — In  this  statement  of  the  law,  the  power  is  supposed  to  be  ap- 
plied to  the  wheel. 

Proportion  :  P  :  W ::r  :  R. 

Note. — The  diameters  or  the  circumferences  may  be  used  instead  of 
the  radii. 

III.  The  elements  involved  in  problems  of  the  inclined  plane 
are  2Jower  (P),  weight  (IF),  the  height  of  the  plane  (/t),  the 
length  of  the  plane  {I),  and  the  base  of  the  plane  (6). 

Laws  of  the  Inclined  Plane  :  ( 1 )  When  the  power  acts 
parallel  to  the  plane,  The  power  is  to  the  weight  as  the  height  of 
the  plane  is  to  its  length.  (2)  When  the  power  acts  horizontally, 
The  power  is  to  the  weight  as  the  height  of  the  plane  is  to  its  base. 

Proportions  :  ( 1 )  P :  W:  :h  :l. 
(2)  P:W::h:b. 

IV.  The  elements  involved  in  problems  of  the  screw  are 
power  (P),  weight  {W),  the  distance  between  the  threads  of 
screw  {d),  and  the  circumference  through  which  the  j)ower 
moves  (C). 

Law  of  the  Screw  :  The  power  is  to  the  weight  as  the  distance 
between  the  threads  is  to  the  circumference  through  which  the  power 
moves. 

Proportion  :  P  :W :  :  d  :  C. 

Note, — These  proportions  are  formulas,  stated  in  the  form  of  propor- 
tion instead  of  that  of  equations. 


ADVANCED    ARITHMETIC.  819 

EXAXFLSS. 

1.  The  pilot-wheel  of  a  boat  is  3  feet  in  diameter;  the  axle, 
6  inches ;  the  resistance  of  the  rudder  180  pounds.  What 
power  applied  to  the  wheel  will  move  the  rudder  ? 

6in.  =  ift. 
Proportion :  (1)  P:  180 : :  ^ :  3. 

.'.  the  power  will  be  30  lb.  to  balance  the  resistance  of  the  rudder; 
anything  more  than  30  lb.  will  move  it. 

2.  The  distance  between  the  threads  of  a  screw  is  1^  inches, 

the  circumference  through  which  the  power  moves  is  18^  feet. 

What  weight  will  a  power  of  20  pounds  support  (neglecting 

friction)? 

13Ht.  =  162in. 

Proportion  :  (1)  20:  W::  U :  162. 

-      ^  (2)  TF=  20x162^2160. 

.-.  the  required  weight  is  2160  lb. 

3.  A  man  exerts  120  pounds  of  power  in  supporting  a280-lb. 
barrel  upon  a  7-ft.  plane  extending  from  the  ground  to  a 
wagon :  how  high  is  the  wagon  (pressure  exerted  in  the  direc- 
tion of  plane)? 

Proportion:  (1)  120 :  280 : /i :  7. 

^^^''-    280    -^• 
.*.  the  height  is  3  ft. 

4..  The  length  of  a  lever  is  16  feet;  weight  at  one  end,  100 
pounds  :  what  power  must  be  put  at  the  other,  4  feet  from  the 
fulcrum,  to  support  the  weight  ? 

Proportion  :  (1)  P:  100 : :  12 : 4. 

(2)  P=i5^  =  300. 
.'.  the  required  power  is  300  lb. 


320  ADVANCED    ARITHMETIC. 

EXEKOISE  CXXXII. 

Name  the  elements  and  give  the  law  of — 

1.  The  lever. 

2.  The  wheel  and  axle. 

3.  The  inclined  plane. 
If.  The  screw. 

5.  State  the  formula  (proportion)  for  each. 

6.  Suppose  a  power  of  75  pounds  be  applied  to  one  end  of  a 
12-foot  lever  to  support  a  load  at  the  other  end :  what  will  be 
the  load  when  the  fulcrum  is  at  the  center  ?  when  the  fulcrum 
is  3  feet  from  the  weight  ? 

7.  In  a  nut-cracker  the  nut  is  placed  1  inch  from  the  hinge 
(fulcrum),  the  hand  6  inches.  If  I  exert  a  pressure  of  10 
pounds,  how  many  pounds  of  resistance  does  the  nut  furnish  ? 

8.  Four  men  hoist  an  anchor  weighing  1  ton ;.  the  barrel  of 
the  capstan  is  8  inches  in  diameter ;  the  circle  described  by  the 
handspikes  is  6  ft.  8  in.  in  diameter.  How  great  a  force  mUst 
each  man  exert  ? 

9.  A  power  of  70  pounds  on  a  wheel  10  feet  in  diameter  bal- 
ances a  weight  of  800  pounds :  find  the  diameter  of  the  axle. 

10.  The  base  of  an  inclined  plane  is  10  feet,  the  height  3 
feet.  What  force  applied  parallel  to  the  base  will  balance  a 
weight  of  2  tons  ? 

11.  In  No.  10,  if  the  force  applied  be  500  pounds,  what 
weight  will  it  support  ? 

12.  A  weight  of  800  pounds  rests  on  an  inclined  plane  8  feet 
high,  being  held  in  equilibrium  by  a  force  of  25  pounds,  acting 
parallel  with  the  plane  :  find  the  length  of  the  plane. 

13.  How  great  a  pressure  will  be  exerted  by  a  power  of  15 
pounds  applied  to  a  screw  whose  head  is  1  inch  in  circumfer- 
ence, and  whose  threads  are  -J  inch  apart  ? 


ADVANCED   ARITHMETIC.  321 

D.    MISCELLANEOUS. 

154.  Partitive  Proportion.  —  Partitive  Pro- 
portion is  the  process  of  dividing  a  given  number  into  parts 
having  a  given  ratio  to  one  another. 

EXAMPIiES. 

1.  .Divide  $25  into  two  parts,  in  the  ratio  of  2  to  3. 

Note. — If  the  two  answers  were  obtained  and  the  first  answer  divided 
into  two  parts  and  the  second  answer  into  3  parts,  there  would  then  be 
5  parts  all  of  the  same  size.  These  5  parts  then  represent  the  amount  to 
be  divided,  or  $25;  2  parts,  the  first  answer;  and  3  parts  the  second  an- 
swer.   Hence, 

Proportions :  (1)  5  parts  :  2  parts  :!  $25  :  $(  )? 
(2)  5  parts  :  3  parts  : :  $25  :  $(  )  ? 
.  •.  the  answers  are  $10  and  $15. 

2.  A  lever  33  feet  long  has  a  power  of  50  pounds  at  one  end 
and  a  weight  of  60  pounds  at  the  other  end.  Where  is  the  ful- 
crum, if  the  power  balances  the  weight? 

From  the  law  of  the  lever,  we  know  that  — 

50  lb. :  60  lb. : :  wt.  arm  :  pr.  arm. 

Then  the  weight  arm  must  be  to  the  power  arm  as  50  to  60,  or  5  to  6. 
Now,  the  problem  is  to  divide  33  feet  into  2  parts  in  the  ratio  of  5  to  6. 

Proportions  :  (1)  11  parts  :  5  parts  ::  33  ft. :  (  )? 
(2)  11  parts  :  6  parts  : :  33  ft. :  {  )  ? 
.*.  the  arms  are  15  ft.  and  18  ft. 

3.  Divide  65  bushels  of  wheat  into  3  parts,  in  the  ratio  of  3 
to  2  to  1^. 

3pt.+2pt.+l^pt.  =  6|  pt. 

Proportions  :  (1)  Ql  parts  :  3  parts  : :  65  bu. :  (  )  bu.? 

(2)  6^  parts  :  2  parts  : :  65  bu. :  (  )  bu.? 

(3)  6^  parts  :  U  parts : :  65  bu.  t  (  )  bu.? 

.-.  thereqd.  partsare30bu.,20bu.,and  15  bu. 


322  ADVANCED    ARITHMETIC. 

EXERCISE  CXXXIII. 

1.  Divide  $750  into  2  parts  in  the  ratio  of  1  to  4;  in  the 
ratio  of  4  to  11. 

2.  Divide  $220  in  the  ratio  of  3  to  7  to  12. 

8.  A  and  B  balance  on  opposite  ends  of  a  pole  10  feet  long. 
A  weighs  120  pounds,  and  B  180  pounds  :  where  is  the  fulcrum? 

4.  At  the  opposite  ends  of  a  lever  20  feet  long,  two  forces  are 
acting  whose  sum  is  1200  pounds.  The  two  arms  of  the  lever 
are  as  2  to  3.  What  are  the  two  forces  when  the  lever  is  in 
equilibrium  ? 

6.  A  fails,  and  can  pay  his  three  creditors,  E,  F,  and  G,  but 
$750.  He  owes  E  $700,  F  $500,  and  G  $300:  how  much  should 
each  receive  ? 

155.  Mixture  Problems. — 

1.  Different  kinds  of  tea  sold  as  follows :  10  lb.  for  $7,  30  lb. 
for  $18,  and  20  lb,  for  $8.     Find  the  average  price  per  pound. 

Solution  :  (1)  Price  of  10  lb.  =  $7. 

(2)  Price  of  30  lb.  =  $18. 

(3)  Price  of  20  lb.  =  $8. 
(l)+(2)+(3)  =  (4)  Price  of  60  lb.  =$33. 

eV  of  (4)  =  (5)  Price  of    1  lb.  =  $.55. 

2,  A  mixture  composed  of  sugar  at  2/  per  lb.  and  sugar  at 
5/  per  lb.  is  worth  3/  per  lb.     How  is  it  mixed  ? 

Note. — There  is  a  gain  on  the  2^  sugar  when  it  becomes  a  part  of  a 
mixture  valued  at  3^  per  lb.,  and  a  loss  on  the  bf  sugar.  They  must  be 
mixed  in  such  a  ratio  as  will  make  the  gain  on  the  one  equal  to  the  loss 
on  the  other. 

Solution  :  (1)  Gain  on  1  lb.  of  2f  sugar  at  3^  =  If. 
(2)  Loss  on  1  lb.  of  5^  sugar  at  3^  =  2^. 
N  (2)  =  (3)  Amt.  of  5f  sugar  reqd.  to  lose  2^  =  1  lb. 

i  of  (3)  =  (4)  Amt.  of  5<*  sugar  reqd.  to  lose  If  =  ^  lb. 
(1)  =  (5)  Amt.  of  2f  sugar  reqd.  to  gain  lf  =  l  lb. 

Comparing  (5)  and  (4),  we  see  that  the  mixture  must  be  made  in  the 
ratio  of  1  lb.  of  the  2f  sugar  to  |  lb.  of  the  5f  sugar,  or  in  the  ratio  of  2  to  1. 


ADVANCED   ARITHMETIC.  828 

3.  How  many  pounds  of  coffee  at  25/  per  lb.  must  I  mix 
with  15  lb.  at  80/  per  lb.  to  make  a  mixture  worth  28/  ? 

Solution  :  (1)  Loss  on  1  lb.  of  30^  coffee  at  28^  =  2^. 
15x(l)  =  (2)  Loss  on  15  lb.  of  30<*  coffee  at  28<^  =  30^,  whole  loss. 
(3)  Gain  on  1  lb.  of  25,^  coffee  at  28^  =  3^. 
(3)  =  (4)  Amt.  of  25^  coffee  reqd.  to  gain  3<*  =  1  lb. 
10x(4)  =  (5)  Amt.  of  25^  coffee  reqd.  to  gain  30,^  =  10  lb.,  answer. 

4-.  Mix  two  kinds  of  coffee  at  20/  and  25/  per  lb.,  to  make  a 
mixture  of  25  lb.  worth  24/  per  lb. 

Solution  :  (1)  Gain  on  1  lb.  of  20^  coffee  at  24^  =  4^. 
(2)  Loss  on  1  lb.  of  25<^  coffee  at  24^  =  1^. 

(1)  =  (3)  Amt.  of  20^  coffee  reqd.  to  gain  4^  =  1  lb. 
i  of  (3)  =  (4)  Amt.  of  20<5  coffee  reqd.  to  gain  l^  =  i  lb. 

(2)  =  (5)  Amt.  of  25^  coffee  reqd.  to  lose  1^  =  1  lb. 

From  (4)  and  (5),  we  see  the  mixture  must  be  in  the  ratio  of  i  lb.  of 
20f  coffee  to  1  lb.  of  the  25<^  coffee,  or  the  ratio  of  1  to  4.  The  problem 
now  is,  to  divide  25  lb.  in  the  ratio  of  1  to  4. 


(6)  5  parts :  1  part   : :  25  lb. :  (  )  ? 

(7)  5  parts  :  4  parts  t :  25  lb. :  (  )  ? 

.a      n  4-     1 X  25  lb.     -. -, 

.  .  Smaller  part  = =  5  lb. 

o 

and  the  larger  part  = '  =  20  lb. 

5 


EXERCISE  CXXXIV. 

1.  How  many  pounds  of  sugar  at  8/  per  lb.  must  I  mix  with 
10  lb.  at  11/  per  lb.  to  make  a  mixture  worth  10/  per  lb.? 

2.  How  must  I  mix  coffee  at  24/  per  lb.  with  coffee  at  80/ 
per  lb.  to  make  a  mixture  worth  26/  per  lb.? 

3.  I  mix  sugar  at  9/  per  lb!  with  72  lb.  at  5/;  the  whole 
mixture  is  worth  7/  per  lb. :  how  much  of  the  9/  sugar  did  I 
use? 

4-.  A  vintner  mixed  two  kinds  of  wine,  one  at  40/  per  pint 
and  the  other  at  60^  per  pint.  He  finds  that  the  mixture  is 
worth  just  55/  per  pint :  how  did  he  mix  it  ? 


324  ADVANCED   ARITHMETIC.  ^ 

5.  A  saloon-keeper  watered  his  60/  wine,  and  sold  it  at  45/ 
per  pint :  if  there  were  6  gallons  and  1  quart  of  water  added, 
how  much  wine  was  there  ? 

6.  A  man  mixed  coffee  at  26/  per  lb.  with  coffee  at  82/  per 
lb.,  making  a  mixture  of  48  lb.  worth  80/  per  lb. :  how  much 
of  each  did  he  use  ? 

7.  I  mix  wine  at  40/  and  50/  per  pint  with  wine  at  60/  per 
pint;  the  50/ and  60/  wine  is  mixed  in  the  ratio  of  2  of  the  50 
to  1  of  the  60 :  if  the  mixture  is  worth  48/  per  pint,  how  do  I 
mix  it  ? 

8.  If  quarters  and  dimes  are  mixed  in  such  a  ratio  that  the 
same  money  might  be  molded  into  the  same  number  of  15/ 
pieces,  how  are  they  mixed  ? 

156.  Equation  of    Payments.— Equation  of 

Payments  is  the  process  of  finding  the  date  on  which  one 

\         payment  may  be  made  in  settlement  of  several  debts  or  an  ac- 

*         count  of  several  items,  falling  due  at  different  times,  without 

loss  to  either  debtor  or  creditor. 

EXAMPLES. 

1.  I  owe  $600  due  in  10  months,  $800  due  in  9  months,  and 
$1000  due  in  6  months,  without  interest.  When  can  I  settle 
all  together  ? 

Solution :  Suppose  that  I  should  settle  all  now,  then  I  would  lose  — 

(1)  Use  of   $600  for  10  mo.  =  use  of  $1  for  6000  mo. 

(2)  Use  of    $800  for   9  mo.  =  use  of  $1  for  7200  mo. 

(3)  Use  of  $1000  for    6  mo.  =  use  of  $1  for  6000  mo. 

$2400,  amt.  paid.    Use  of  $1  for  19200  mo. ,  loss. 

(4)  Use  of  $1  for  19200  mo.  =  use  of  $2400  for  8  mo. 
.'.  8  months  from  now,  $2400  will  settle  the  debt. 

2.  My  account  with  Jones  is  as  follows :  I  owe  him  $600  due 
in  11  months,  $800  due  in  4  months,  $500  due  in  1  year,  and 
$200  due  now ;  he  owes  me  $700  due  in  8  months  and  $600  due 


ADVANCED    ARITHMETIC.  825 

in  9  months.     When  will  $800  settle  the  account  ?    and  how 
much  will  be  due  1  year  hence,  interest  being  6%? 

Solution :  Assume  now  as  the  date  of  settlement. 

(1)  Use  of  $600  for  11  mo.  =  use  of  $1  for  6600  mo. 

(2)  Use  of  $800  for   4  mo.  =  use  of  $1  for  3200  mo. 

(3)  Use  of  $500  for  12  mo.  =  use  of  $1  for  6000  mo. 

(4)  $200  due  now. 


$2100,  amt.  paid.    Use  of  $1  for  15800  mo.,  loss. 

(5)  Use  of  $700  for  8  mo.  =  use  of  $1  for  5600  mo. 

(6)  Use  of  $600  for  9  mo.  =  use  of  $1  for  5400  mo. 

$1300,  amt.recd.    Use  of  $1  for  11000  mo.,  gain. 

(7)  $2100 -$1300  =  $800,  balance. 

(8)  Use  of  $1  for  15800  mo.  -use  of  $1  for  11000  mo.  =  use 

of  $1  for  4800  mo.,  net  loss. 

(9)  Use  of  $1  for  4800  mo.  =  use  of  $800  for  6  mo. 

.*.   the  time  of  payment  is  6  mo.  from  now.    (istans.) 
(10)  Amt.  of  $800  at  Q%  for  |  yr.  =  $24+ $800  =  $824.     (2d  ane.) 

3.  I  bought  goods  Jan.  1, 1898,  on  80  days'  credit,  $600 ;  Feb.  , 

10,  on  60  days'  credit,  $5(X).     When  can  I  settle  the  whole  by  < 

paying  $1100? 

Solution:  Assume  Jan.  1st  as  the  date  of  settlement.  The  first  debt 
will  be  paid  30  days  before  it  is  due,  and  the  second  will  be  paid  100  days 
before  it  is  due. 

(1)  Use  of  $600  for    30  days  =  use  of  $1  for  18000  days. 

(2)  Use  of  $500  for  100  days  =  use  of  $1  for  50000  days. 

$1100,  amt.  paid.     Use  of  $1  for  68000  days,  loss. 

(3)  Use  of  $1  for  68000  days  =  use  of  $1100  for  62  days,  nearly. 
62  days  after  Jan.  1, 1898,  is  March  4, 1898. 

Note.— A  part  of  a  day  is  always  counted  as  a  full  day  in  equating  the 
time  of  payment.  This  is  plain,  for  in  the  above  problem,  as  the  time 
was  more  than  61  days,  it  must  fall  on  the  62d  day. 

These  solutions  are  given  in  full.     After  the  pupil  writes  out 


326  ADVANCED    ARITHMETIC. 

a  few  in  this  way,  he  should  shorten  the  solution.     Thus,  for 

Example  1 : 

Solution:  (1)    600x10  =  6000. 

(2)  800  X  9  =  7200. 

(3)  1000 X  6  =6000. 
2400  19200. 

19200  +  2400  =  8. 
Answer,  $2400,  8  mo.  from  now. 

EXERCISE  CXXXV. 

1.  I  owe  $400  due  in  4  months  and  $800  due  in  7  months, 
each  bearing  interest  at  8%.  Find  the  equated  time  and  the 
amount  due. 

2.  I  owe  Smith  $1200  due  in  8  months  and  $600  due  in  5 
months ;  Smith  owes  me  $2000  due  in  10  months :  when  will 
$200  settle  what  he  owes  me  ? 

3.  A  merchant  owes  his  creditor  for  goods  bought,  as  follows : 
July  3,  $640;  July  15,  $300;  Aug.  10,  $520:  when  will  $1460 
settle  his  account  ? 

4..  If,  in  No.  3,  the  merchant  pays  $500  Aug.  1st,  when  can 
he  settle  the  balance  with  $960  ? 

5.  If,  in  No.  3,  each  amount  is  a  note  due  in  60  days,  bearing 
10%  interest,  find  the  equated  time  of  payment  and  the  amount 
due. 

157.   Partner sliip. 

EXAMPIiES. 

i.  A,  B  and  C  form  a  partnership,  A  puts  in  $200,  B  $300, 
and  C  $400.     They  gain  $450.     Divide  it  among  them. 

Solution  :  The  whole  capital  is  $900.  Each  partner's  capital  is  to  the 
whole  capital  as  his  share  of  the  gain  is  to  the  whole  gain.     Hence  — 

(1)  For  A,  $200 :  $900 : :  $(  ) :  $450  ? 
.-.  A's  gain  is  $100. 

(2)  For  B,  $300 :  $900 : :  $(  ) :  $450  ? 
.*.  B's  gain  is  $150. 

(3)  For  C ,  $400 :  $900 : :  $(  ) :  $450  ? 
.-.  C's  gain  is  $200. 


ADVANCED   ARITHMETIC.  827 

2.  A  invests  $100  for  6  months ;  B,  $200  for  2  months.     They 
gain  $50.     Divide  it  between  them. 
Solution :  "* 

(1)  Earnings  of  $100  for  6  mo.  =  earnings  of  $600  for  1  mo.,  A's  part. 

(2)  Earnings  of  $200  for  2  mo.  =  earnings  of  $400  for  1  mo.,  B's  part. 

(3)  The  whole  earnings  =  earnings  of  $1000  for  1  mo. 

m  For  A    \    Earnings  of    ).    j     earnings  of     }  .  •*/  n -3;.^? 

(4)  J^or  A,  j  ^g(^  j^j.  1  jno.  r    \  $1000  for  1  mo.  T  "^^  ''^^^ 

.'.  A's  gain  is  $30. 

Note. —  The  "  1  mo."  may  be  omitted,  as  it  is  the  same  in  both  the  1st 
and  the  2d  terms,  and  the  proportion  may  be  shortened  as  in  (5). 

(5)  ForB,  $400:  $1000::  $(  ):$50? 

.*.  B's  gain  is  $20. 

EXERCISE  OXXXVI. 

1.  Three  persons  purchase  a  farm  for  $2800,  of  which  A  pays 
$1200,  B  $1000,  and  C  $600.  How  shall  they  divide  the  rent, 
which  pays  $224  a  year  ? 

;g.  A,  B  and  C  owned  a  cargo  of  corn  valued  at  $3475.60.  A 
owned  i,  B  ^,  and  C  the  remainder.  How  much  did  each  lose 
if  the  cargo,  which  was  insured  for  $2512,  was  lost  ? 

S.  A,  B  and  C  form  a  copartnership.  B  puts  in  $250  for  6 
months,  C  $275  for  8  months,  and  A  $450  for  4  months.  Divide 
their  gain,  $825. 

^.  A,  B  and  C  form  a  copartnership  for  1  year,  and  invest 
respectively  $9600,  $8400,  and  $7200.  After  4  months,  A  in- 
vests $2000  additional,  B  $1400,  and  C  $800.  They  gain 
$12800:  what  is  each  man's  gain  ? 

158.  Longitude  and  Time. —  I.  Longitude. 
Longitude  is  the  distance  east  or  west  of  the  prime  meridian 
measured  in  degrees  (°),  minutes  (^),  and  seconds  (^^). 

TABIjE. 

1°  =  60\ 

r  -  60  ^ 


328  ADVANCED    ARITHMETIC. 

The  meridian  of  Greenwich,  England,  has  been  selected  as 
the  prime  meridian.  180°  east  from  the  prime  meridian  (0°) 
reaches  half  around  the  earth.  Any  place  on  that  half  of  the 
earth  is  said  to  be  in  east  longitude.  180°  from  0°  west  reaches 
around  the  other  half  of  the  earth.  Any  place  on  that  half  of 
the  earth  is  said  to  be  in  west  longitude.  The  longitude  of  a 
place  is  its  distance  {less  than  180°)  east  or  ivest  of  the  prime 
meridian.  The  distance  measured  in  one  direction  (east  or 
west)  around  the  earth  is  860°. 

II.  Time. — (1)  Sun  Time:  The  sun  appears  to  travel  west- 
ward around  the  earth  (860°)  once  in  24  hours.  It  passes  over 
15°  of  longitude  in  1  hour  of  time ;  15'  of  longitude  in  1  min- 
ute of  time ;  and  15'^  of  longitude  in  1  second  of  time.  There- 
fore the  difference  of  longitude  between  two  places,  expressed  in 
^  '  ^\  is  numerically  15  times  the  difference  of  time ^  expressed  in 
hours,  minutes,  seconds.  Of  two  places  on  meridians  nearer  to 
each  other  than  180°,  the  place  west  is  earlier  and  the  one  east  is 
later,  except  when  the  I.  D.  L.  (see  below)  and  midnight  are 
both  between  the  two  places ;  in  which  case  the  place  west  is 
later  by  the  difference  between  their  difference  of  time  and  24- 
hours. 

(2)  The  Day :  Every  day  begins  first  at  or  near  the  180th 
meridian.  On  this  account  the  180th  meridian  is  often  called 
the  International  Date  Line  ( I.  D.  L.).*  When  midnight  is  on 
the  I.  D.  L.,  there  is  one,  and  but  one,  day  on  all  the  earth, 
(say)  Sunday.  As  midnight  moves  westward,  it  changes  the 
day  on  every  part  of  the  earth  over  which  it  passes  to  the  new 
day,  Monday.     When  it  has  gone  all  the  way  around  and  again 


*The  International  Date  Line  used  to  have  a  very  winding  course.  After  coming  south- 
ward through  Bering  Strait,  it  took  a  southwesterly  course,  passing  west  of  the  Philippine  Is- 
lands, then  bending  eastward,  then  southward  into  the  Southern  ocean  near  the  180th  meridian. 
But  now,  after  passing  southward  through  Bering  Strait,  it  swerves  a  little  (20°  or  30°)  to  the 
westward.  Returning  to  the  180th  meridian  about  40°  N.,  it  follows  that  meridian  to  a  point 
about  10°  S.  Then  it  swerves  eastward  about  10°  or  15°.  Returning  to  the  180th  meridian 
again  about  50°  S.,  it  continues  southward  on  that  meridian.  This  description  was  taken  from 
a  diagram  given  in  The  Pathfinder  in  1899. 


ADVANCED    ARITHMETIC.  829 

reaches  the  I.  D.  L.,  it  is  Monday  all  over  the  earth.  When 
any  place  is  between  the  I.  D.  L.  and  the  midnight  meridian, 
if  it  is  west  of  the  I.  D.  L.  and  east  of  midnight,  it  is  the  new 
day ;  if  it  is  east  of  the  I.  D.  L.  and  west  of  midnight,  it  is  the 
old  day. 

(S)  Standard  or  Railroad  Time :  There  are  five  standards  of 
railroad  time  in  the  United  States  and  Canada :  Intercolonial, 
Eastern,  Central,  Mountain,  and  Pacific.  Each  covers  a  lon- 
gitude of  15°.  Intercolonial  commences  52J°  W.,  and  the  Pa- 
cific ends  1271°  W.  The  time  throughout  each  standard  is  the 
sun  time  of  its  central  meridian  :  For  Intercolonial,  60th;  for 
Eastern,  75th;  for  Central,  90th;  for  Mountain,  105th;  and 
for  Pacific,  120th. 

Note. — Railroad  companies  vary  the  boundary -lines  of  these  standards 
in  some  places. 

III.  Problems:  There  are  two  classes  of  problems  usually 
given  in  Longitude  and  Time :  ( 1 )  Given  the  longitudes  of  two 
places  and  the  time  at  one  of  them,  to  find  the  time  at  the 
other;  and  (2)  given  the  times  of  two  places  and  the  longitude 
of  one,  to  find  the  longitude  of  the  other. 

EXAMPIiES. 

1.  When  it  is  noon,  Monday,  150°  W.,  what  is  the  time  120° 
E.? 

Direction. — In  drawing  a  figure  to  repre- 
sent a  longitude  problem,  the  following  will 
help  to  make  your  problem  clear:  (1)  Sketch 
a  circular  figure  to  represent  a  parallel  of  lat- 
itude ;  (2)  place  0°  at  the  top ;  (3)  180°  (I.  D.  L.) 
at  the  bottom ;  (4)  locate  approximately  the 
two  places  under  consideration  and  midnight  p^  =i;  jy  l 

(M.).  M.  =  midnight. 


o 


330  ADVANCED    ARITHMETIC. 

Solution:  (1)  180° - 150°  =  30°. 

(2)  180° -120°  =  60°. 

(3)  30°+60°  =  90°,  difference  of  longitude. 
.'.6  hours  =  difference  of  time. 
6  hours  earlier  than  noon  =  6  a.  m. 
150°  W.  is  in  the  old  day  (Monday). 
120° E.  is  in  the  new  day  (Tuesday). 

2.  When  it  is  6  hr.  40  min.  p.  m.,  Friday,  20°  E.,  what  is 
the  time  140°  E.? 

Solution:  (1)  140° -20°  =  120°,  difference  of  longitude. 
.  • .  8  hr.  =  difference  of  time. 
8  hr.  later  than  6  hr.  40  min.  p.  m.  =2  hr.  40 
min.  a.  m.  of  the  following  day,  Saturday. 


3.  When  it  is  7  hr.  20  min.  a.  m.,  Friday,  160°  W.,  on  what 
meridian  is  it  3  hr.  20  min.  a.  m.  Saturday? 

Solution:  (1)  7  hr.  20  min. -3  hr.  20  min.  =  4  hr.,  dif- 
ference of  time. 
.  •.  60°  =  difference  of  longitude. 
Since  3  hr.  20  min.  a.  m.  is  earlier  than  7  hr.  20  min. 
a.  m,,  the  required  place  is  west.    60"  west  of  160°  W. 
is  140° E. ;  and  it  is  Saturday,  because  it  is  in  the  new 
day  (beyond  the  I.  D.  L.). 

4..  When  it  is  noon  at  Greenwich,  0°,  what  is  the  time  at  St. 
Louis  (standard  time)? 

Solution  :  The  time  of  St.  Louis  is  that  of  90°  W.,  or  6  hr.  earlier  than 
noon  =  6  hr.  a.  m.  of  the  same  day. 

5.  When  it  is  3  o'clock  a.  m.  at  Kansas  City,  where  is  it  2 
o'clock  a.m.?     4  o'clock  a.  m.? 

Solution  :  (1)  Kansas  City  is  in  Central  Time ;  it  is  2  o'clock  a.  m.  on  all 
the  standard  west,  or  7^°  each  side  of  105°  W. 
(2)  It  is  4  o'clock  a.  m.  on  all  the  standard  east  of  Central 
Time,  or  71°  each  side  of  75°  W. 


ADVANCED    ARITHMETIC.  331 


EXERCISE  CXXXVII. 


1.  When  it  is  9  o'clock  p.  m.,  Wednesday,  20°  E.,  what  is 
the  time  at  Greenwich  (0°)? 

2.  When  it  is  9  o'clock  p.  m.,  Wednesday,  20°  E.,  what  is 
the  time  at  30°  W.? 

3.  When  it  is  9  o'clock  ]).  m.,  Wednesday,  20°  E.,  what  is 
the  time  at  120°  W.? 

^.  When  it  is  9  o'clock  p.  m.,  Wednesday,  20°  E.,  what  is 
the  time  at  80°  E.? 

5.  When  it  is  noon  at  Greenwich,  at  what  i:)lace  is  it  2 
o'clock  a.  m.? 

6.  When  it  is  noon  at  Greenwich,  at  what  place  is  it  7 
o'clock  a.  m.? 

7.  When  it  is  noon  at  Greenwich,  at  what  place  is  it  3 
o'clock  p.  m.? 

8.  When  it  is  noon  at  Greenwich,  at  what  place  is  it  8 
o'clock  p.  m.? 

9.  When  it  is  1  hr.  20  min.  p.  m.,  Saturday,  110°  E.,  what 
is  the  time  170°  E.? 

10.  When  it  is  1  hr.  20  min.  p.  m.,  Saturday,  110°  E.,  what 
is  the  time  105°  W.? 

11.  When  it  is  1  hr.  20  min.  p.  m.,  Saturday,  110°  E.,  at  what 
place  is  it  9  hr.  p.  m.? 

12.  When  it  is  1  hr.  20  min.  p.  m.,  Saturday,  110°  E.,  at  what 
place  is  it  11  hr.  p.  m.? 

13.  When  it  is  1  hr.  20  min.  p.  m.,  Saturday,  110°  E.,  what  is 
the  time  at  Boston  ?  at  New  York  ?  at  St.  Louis  ?  at  Cincin- 
nati ?  at  Kansas  City?  at  San  Francisco  (standard  time)? 

IJ^.  When  it  is  noon"  on  the  180th  meridian,  it  is  6  hr.  52 
min.  40  sec.  p.  m.  at  Harrisburg,  Pa.  (sun  time) .  Find  the  lon- 
gitude of  Harrisburg. 

15.  A  watch  is  set  right  at  Pekin,  China,  longitude  116°  26'' 
E.  On  what  meridian  is  the  watch  when,  at  noon,  it  shows  9 
hr.  20  min.  a.m.? 


882  ADVANCED   ARITHMETIC. 

16.  When  it  is  5  minutes  after  4  o'clock  on  Sunday  morning 
at  Honolulu,  longitude  157°  52^  W.,  what  is  the  time  at  Sydney, 
Australia,  longitude  151°  11'  E.? 

17.  The  battle  of  Manila  began  41  min.  after  5  o'clock,  Sun- 
day morning,  May  1.  Manila  is  in  longitude  121°  20'  E.  Wash- 
ington is  in  longitude  77°  W.  What  was  the  time  at  Washington 
when  the  battle  began,  by  sun  time  ?  by  standard  time  ? 

18.  When  it  is  10  o'clock  p.  m.  Wednesday,  20°  E.,  what  is 
the  time  at  Sydney,  Australia  ?  at  Honolulu  ?  railroad  time  at 
San  Francisco,  Cal.?  at  St.  Louis,  Mo.?  at  Boston,  Mass.? 

Note. — When  time  is  required,  always  give  the  day  as  well  as  the  hour 
of  the  day. 


PART  III. 

I.    STUDY  OF  NUMBERS. 
A.     INTEGRAL   NUMBERS. 

159.    Positive    and    Negative    Numbers. — The 

signs  phis  (-f-)  and  minus  (  — )  are  placed  before  numbers, 
when  written  with  figures  or  letters,  to  indicate  opposite  mean- 
ings or  conditions.     Thus, 

If  +$4  means  $4  gain,  -$4  means  $4  loss. 
If  +$4  means  $4  capital,  -$4  means  $4  debt. 
If  +4  means  4  to  he  added,  -4  means  4  to  he  suhtracted. 
If  +4°  means  4°(temperature)  ahove  zero,  -4°  means  4°  heloiu  zero. 
If  +4  ft.  means  4  ft.  traveled  to  the  right, -4  ft.  means  4  ft.  trav- 
eled to  the  left. 

A  Positive  Number  is  a  number  which,  when  written, 
is  preceded  by  the  plus  sign,  expressed  or  implied. 

Note. —  For  convenience,  a  positive  number,  standing  alone  or  first  in 
an  expression,  is  usually  written  without  the  sign,  +. 

A  Neg-ative  Number  is  a  number  which,  when  written, 
is  preceded  by  the  minus  sign. 

Note. — The  sign,  -,  is  never  omitted  before  negative  numbers. 

All  numbers  are  eiiher  positive  or  negative.     To  illustrate: 

5,4,8,2,1,0,  -1*,  -2,  -3,  -4,  -5. 

The  numbers  in  this  series  decrease  in  absolute  value  from 
left  to  right;  each  number,  beginning  with  4,  is  one  less  than 


*  Bead,  minus  one,  minus  two,  and  so  on. 

(333) 


884  ADVANCED    ARITHMETIC. 

the  number  to  the  left  of  it.  Notice  that  from  0  to  the  right 
the  absolute  value  decreases  as  the  numerical  value  increases ; 
—5  is  less  than  —4,  —4  is  less  than  —8,  etc. 

The  question  is  sometimes  asked,  "Can  a  number  be  less 
than  nothing?"  Yes:  a  number  may  be  less  than  nothing. 
Suppose  that  I  have  but  $5  and  owe  nothing,  and  you  ask  me 
how  much  I  am  worth:  I  would  say,  "$5."  That  means  that 
I  am  worth  $5  more  than  nothing,  +$5.  Suppose  now,  that  I 
have  the  $5,  and  in  addition  I  owe  some  one  $8.  I  am  now 
worth  $8  less  than  before,  or  but  $2.  Suppose  again  that  I  owe 
another  debt  of  $3 :  my  worth  is  again  decreased  $8,  and  I  am 
now  worth  $1  less  than  nothing,  —  $1.  If  I  should  have  another 
debt  of  $8,  I  would  then  be  worth  $4  less  than  nothing,  —  $4. 

Note.— It  does  not  follow  that  all  negative  numbers  represent  value 
less  than  nothing.  A  negative  distance  is  not  less  than  nothing;  -10° 
( temperature)  is  not  less  than  nothing. 

160.  Numbers  Expressed  by  Means  of  Let- 
ters.—  In  expressing  numbers  by  means  of  letters,  or  letters 
and  figures,  there  are  four  things  to  be  considered :  (1)  the 
literal  part,  (2)  the  coefficient,  (3)  the  exponent,  and  (4)  the 
sign. 

The  Literal  Part  is  the  letter  or  letters  used  as  factors  in  ex- 
pressing a  number.     Thus, 

In  -5a^,  a  is  the  literal  part ;  in  5a%,  a  and  b  form  the  literal  part. 

The  Coefficient  is  a  factor  (expressed  by  figures  in  this  book), 
placed  on  the  left  of  the  literal  part.     Thus, 

In  Wax,  15  is  the  coefficient. 

Note. —  In  such  expressions  as  ax,  or  -ax^,  the  coeflRcient  is  not  ex- 
pressed, but  is  understood  to  be  1. 

An  Exponent  is  a  small  figure  (sometimes  a  letter),  placed 


ADVANCED    ARITHMETIC.  335 

just  above  and  to  the  right  of  a  factor  to  show  what  power  of 
the  factor  is  used.     (See  p.  179.) 

The  Sign  of  a  number  is  used  to  indicate  whether  the  number 
is  positive  or  negative. 

A  number  of  one  term  is  called  a  Monomial.  A  number  of 
more  than  one  term  is  called  a  Polynomial. 

A  polynomial  of  two  terms  is  called  a  Binomial;  of  three 
terms,  a  Trinomial. 

EXERCISE  CXXXVIII. 

1.  Give  use  of  the  signs  plus  and  minus. 

2.  What  is  a  positive  number  f 

3.  What  is  a  negative  number  ? 

If.  Explain  how  some  numbers  do  have  values  Zess  than  nothing. 

5.  How  many  things  are  to  be  considered  in  expressing  num- 
bers by  letters  or  by  letters  and  figures  ? 

6.  Explain  what  is  meant  by  literal  part.     Give  example. 

7.  Define  coefficient.     Give  example. 

8.  Define  exponent.     Give  example. 

9.  What  does  the  sign  of  a  number  show  ? 

10.  Define  monomial.     Give  example. 

11.  Define  polynomial.     Give  example. 

12.  Define  binomial.     Give  example. 

13.  Define  trinomial.     Give  example. 

161.  Addition. —  Since  we  now  have  to  consider  both 
positive  and  negative  numbers,  our  process  of  addition  must 
be  extended  to  cover  the  addition  of  negative  as  well  as  positive 
numbers. 

In  explaining  the  following  examples,  let  us  assume  that  + 
means  capital  and  —  means  debt. 


ADVANCED    ARITHMETIC. 
BXAMPIiBS. 

1.  Add  +$8  and  +$4. 

Proce.:  +,8+(+»4)= +.12.  result.  J^^^^ 

2.  Add  -$8  and  -$4.  $12  (capital). 

d.o    /    *.s        ^.^         ,  H  (debt)  added  to 

Process:  -$8  +  (-$4)=  -$12,result.  ^g  (debt)  makes  $12 

3.  Add  +$8  and  -$4.  (debt). 

*o    /    *.v        ^.  ,  $4  (debt)  added  to 

Process:  +$8  +  (-$4)= +$4,  result.  ^g  (^^p.^^^^  ^.^^^  ,^^ 

4.  Add  -$8  and  +$4.  (capital). 

$4  (capital)  added 
Process:   -$8+(  +  $4)=  -$4,  result.  to  $8  (debt)  gives  $4 

(debt). 
Let  us  write  these  processes  abstractly  and  compare : 

(1)  +8  +  (+4)=+12. 

(2)  -8  +  (-4)=-12. 

(3)  +8  +  (-4)=+4. 

(4)  -8  +  (  +  4)=-4. 

Observe  that  in  (1)  the  signs  were  alike  (both  plus),  and  we  added  8 
and  4  ;  in  (2)  the  signs  were  alike  (both  minus),  and  we  added  8  and  4 ;  in 
(3)  and  (4)  the  signs  were  unlike  (one  plus,  the  other  minus),  and  we  sub- 
tracted 4  from  8.  In  every  case,  the  sign  of  the  result  is  the  same  as  that 
of  the  larger  addend. 

Rule  for  Addition  :  When  the  signs  are  alike,  add  the  num- 
bers; lohen  unlike,  subtract  the  smaller  from  the  larger,  giving  the 
result  the  sign  of  the  larger  addend. 

5.  Add  50,  +38,  -14,  -78,  -52,  +28,  +44.  . 

I.  Process:  (1)  50+38  =  88  ;  (2)  88-14  =  74  ;  (3)  74-73  =  1 ;  (4)  1-52= -51 ; 
(5)  -51  +  23= -28;  (6)  -28  +  44  =  16,  result. 

Note.— The  process  may  be  much  shortened  by  uniting  the  positive 
numbers  into  one  amount  and  the  negative  numbers  into  another.  The 
difference  of  these  amounts  preceded  by  the  sign  of  the  greater  will  be 
the  mathematical  sum.    Thus: 

II.  Process:  (1)  50+38+23+44  =  155. 

(2)  -14-73-52=  -139. 

(3)  155-139  =  16,  result. 


ADVANCED    ARITHMETIC.  887 

Literal  numbers  are  similar,  when  they  are  composed  of  the 
same  letters  affected  by  the  same  exponents.     Thus : 

ah^  and  25  aW  are  similar,  but 
a%  and  ah"^  are  dissimilar. 

Principles:  1.  Only  similar  numbers  can  he  added  to  form  one 
term. 

2.  The  addition  of  dissimilar  numbers  is  indicated  by  placing 
between  them  the  proper  sign. 


6. 

7. 

8, 

-4  a 

6xy 

7x^y 

-7a 

4:xy 

-6xhj 

+5  a 

-dxy 

-Ux'y 

+15  a 

-Sxy 

-{-llx'y 

—  a 

result. 

-Sxy 
-11  xy, 

result. 

-hx'^y 

+8  a, 

0, 

0,  result. 

9.  Add  -76,  +96,  -\-6xy,  +46,  -7xy-j-z. 

Process  :   -7b^6xy-\-z 
+Qb-7xy 
+46 

6b -xy+z,  result. 

10.  Add  12xy'^-lbax-c,  7aa;+9c,  -bxif+17 ax-12c. 

Process  :  12 xy^ -lb ax     -  c 
+  7  ax  +9c 
-5xy^-i-17ax-12c 
7xy^+9ax-4c,  result. 

11.  Add5(x-2/),  -7ix-y),  l^x-y). 

Process  :      5  (.r  -  y) 
-7{x-y) 
15{x-y) 
ld{x-y),  result. 

Note. — When  the  expressions  in  parentheses  are  similar,  they  may  be 
treated  as  one  number. 

As,  bxNo. -7y  No. +  15xNo.  =  13xNo. 


338  ADVANCED    ARITHMETIC. 

EXERCISE  CXXXIX. 

1.  (jiwe  rule  for  addition.     Give  principles. 

2.  Explain  examples  1  to  4,  letting  the  numbers  represent 
distance, —  the  positive  numbers  distance  measured  to  the 
right,  and  negative  numbers  distance  measured  to  the  left. 

Thus,  In  Example  1, 

+  8,  to  the  right. 
+4,  to  the  right. 
+  12 ;  or,  12  to  the  right  of  starting-point. 

In  Example  3, 

+  8,  to  the  right. 

-4,  to  the  left. 

+4 ;  or,  still  4  to  the  right  of  starting-point. 


Add  (orally) : 

3. 

4- 

6. 

6. 

7. 

-7 

+7 

+7 

-7 

14 

-5 

+5 

-5 

+5 

-9 

8. 

9. 

10. 

11. 

12. 

-5 

+9 

-17 

+4 

-13 

+7 

-6 

+4 

-18 

+  12 

-4 

+3 

_2 

+5 

-6 

IS. 

u. 

15. 

16. 

17. 

—  a 

-6x 

-{-12  xy 

-3  A 

-20  xY 

+5  a 

+25  a: 

-16xy 

+5  A 

+  18a:y 

-7a 

-3a; 

-\-9xy 

-12a^x 

+4a:y 

+  a 

-11a: 

-12  xy 

+4a2a; 

-14a;Y 

18, 

19. 

20. 

7{x- 

■22/) 

-Hx^-y) 

9(x+2/)^ 

-9{x- 

■22/) 

+Hx'-y) 

- 

-Hx-^yY 

-{-(X- 

-22/) 

-\2(x^-y) 

^(x^-yy 

ADVANCED    ARITHMETIC.  839 

Copy  and  add  : 

21.  12a-9c,  -4a+17c,  -7  c+15a,  27ft-c. 

22.  42a^-5ab,  -SQ  a^-\-12ab,  7a^-15ab,  -ab-14:a\ 

23.  a'-2abW.  5a2-9a6+4?>2,  -12a2+26a6-12  6^,  36a-^ 
-24  62+7  a6. 

21^.  \la%-^nab''-\-a%\  ba'b -21  ab^-^^a%\  lA.aW-bd'W^ 
9a^b,  -4a262+5a62+3a26. 

25.  9xhj-{-4:X-\-Sij-9y\  4:y^-2x^y-5x^Sy,  12x-14:y-Qy' 
-2x^y,  x^y-x+y-y^. 

162.  Subtraction. —  It  will  help  to  explain  the  process 
of  subtracting  positive  and  negative  numbers,  if  we  will  note 
the  fact,  that  any  monomial  may  be  expressed  as  a  binomial ,  one  of 
whose  terms  is  positive  and  the  other  negative,  and  that  the  numeri- 
cal value  of  either  term  may  be  increased  at  will,  provided  the  nu- 
merical value  of  the  other  be  correspondingly  increased. 

EXAMPLES. 

1.  Represent  +5  by  an  expression  containing  +8,  +12,  +17, 
-5,  -14,  -21. 

(1)  5=  8-  3.  (4)  5  =  10-  5. 

(2)  5  =  12-  7.  (5)  5  =  19-14. 

(3)  5  =  17-12.  (6)  5  =  26-21. 

2.  Represent  —7  by  an  expression  containing  +4,  +12,  +15, 
-9,  -13,  -25. 

(1)  -7= -11+  4.  (4)  -7=-  9+  2. 

(2)  -7= -19  +  12.  (5)  -7= -13+  6. 

(3)  -7= -22+15.  (6)  -7= -25  +  18. 

The  process  of  subtraction  consists  in  taking  from  the  minu- 
end the  number  expressed  by  the  subtrahend. 

KXAMPIiES. 

3.  From  +8  take  +4. 

(1)  +8 -(+4)=  +4,  result.    Taking  +4  from  +8  leaves  +4. 


340  ADVANCED    ARITHMETIC. 

4.  From  -8  take  -|-4. 

(1)  -8= -12+4.  The  -8  is  the  same  as  -12  +  4;  tak- 

(2)  (-12+4) -(+4)=  -12,  result.       ing  away  the  +4  leaves  -12. 

5.  From  +8  take  —4. 

(1)  +8=  +12-4.  The  +8  is  the  same  as  +12-4;  tak- 

(2)  (+12- 4) -(-4)=  +12,  result.        ing  away  the  -4  leaves  +12. 

6.  From  —8  take  —4. 

(1)  -8-(-4)= -4,  result.    Taking  -4  from  -8  leaves  -4. 

Note. — The  same  result  will  be  obtained  by  the  capital  and  debt  plan 
of  explanation.     (See  Addition,  p.  336.) 
Let  us  compare  these  examples : 

(1)  +8-(+4)=+  4. 

(2)  -8-(+4)=-12. 

(3)  +8-(-4)=+12. 

(4)  -8-(-4)=-  4. 

Observe  that,  when  the  signs  of  the  subtrahend  and  minuend  w^ere 
alike  (in  (1)  and  (4)),  we  took  the  difference  between  the  8  and  the  4; 
when  the  signs  of  the  subtrahend  and  minuend  were  unlike  (in  (2)  and 
(3) ),  we  took  the  sum  of  the  8  and  the  4. 

As  we  might  have  expected,  this  is  exactly  the  reverse  of  our  com- 
parison in  addition. 

Suppose  that  in  the  four  examples  above,  we  change  the  sign 
of  each  subtrahend  and  add  (instead  of  subtracting)  and  see 

what  results  we  get : 

(1)  +8+(-4)=+4. 

(2)  -8+(-4)=-12. 

(3)  +8  +  (+4)=+12. 

(4)  -8+(+4)=-4. 

These  results  are  just  the  same  as  above,  and  we  have  sub- 
tracted by  changing  the  signs  of  the  subtrahends  and  adding. 

Rule  for  Subtraction  :  Conceive  the  sign  of  the  subtrahend 
to  he  changed,  {the  -\-to  —  or  the  —  to  -\-) .  and  proceed  as  in  addi- 
tion. 

Note. — Do  not  actually  change  the  sign  of  the  subtrahend  on  paper 


ADVANCED    ARITHMETIC.  341 

or  slate  in  written  work,  but  do  it  mentally.     An  actual  change  of  written 
signs  leads  to  confusion  in  reviewing  the  work. 

Principles:  1.  The  result  can  he  expressed  in  one  term  only 
when  the  minuend  and  subtrahend  are  similar. 

2.  The  subtraction  of  dissimilar  numbers  is  indicated  by  placing 
between  them  the  proper  sign. 

7.  From  12-5+4  take  8-8+2. 

(1)  12-5+4  =  11.  Collect  the  terms  of  both  minuend  and 

(2)  8-3+2=  7.         subtrahend;  change  the  sign  of  the  sub- 

(3)  11-7  =  4.  trahend,  and  proceed  as  in  addition.     Or, 
^           i2_t5+4-84-S-2  =  4        change  all  signs  of  the  subtrahend, 

'  *       and  then  collect. 

8.  5+17-3-(4+8-10+2)  =  ? 

Process:  5+ 17 -3 -(4+8- 10+2)  = 

5  +  17-3-  4-8  +  10-2  =15,  result. 

Note. — The  -before  the  parentheses  shows  that  all  the  numbers  within 
are  to  be  subtracted,  and  that  the  expression  within  the  parentheses  is 
a  subtrahend. 

Rule  for  Clearing  of  Parentheses  Preceded  by  a  Minus 
Sign:  Drop  the  parentheses  with  the  minus  sign,  and  change  the 
sign  of  every  term  within. 

9.  10,  11.                          12 

7  a  -xy  +3a;2  -17xy' 

5  a  +5.^2/  —9x^  —^xy"^ 

2  a,  result.  —6  a;?/,  result.  +12  .t^,  result.  —  12  a:?/^,  result. 

13.  7x^-\-4:ax-\-2b^-{-bb^^7xa-5x^)  =  {   )? 

Process :        7  x^-\-4:ax+2b^ 
-  5x^+7 a.r-5b^ 
12a;2-8ax+7  62,  result. 
Or, 

7x^-{-4ax+2h'^-{-bh^  +  7ax-5a^)  = 
7  a;2+ 4  ax +2b^  +  bb^-7  ax + 5  x^  = 

12x^-3  ax +71/,  result. 


342  ADVANCED    ARITHMETIC. 

1^.  From5(a4-:c)  take  2(a+a;). 

5(a+a:) 

3(a  +  ar),  result. 

EXERCISE  CXL. 

1.  Explain  examples  3  to  6,  letting  positive  numbers  mean 
capital  and  negative  numbers  mean  debt;  letting  positive  num- 
bers mean  distance  moved  to  the  right,  and  negative  numbers 
distance  moved  to  the  left. 

2.  Give  the  rule  for  subtraction. 

3.  Give  the  'principles. 

J/..  Give  the  rule  for  removing  parentheses  preceded  by  a  minus 
sign. 

5.  Do  you  change  any  signs  in  removing  parentheses  pre- 
ceded by  a  plus  sign  ? 


Subtract  (ora^ 

lly)'. 

6. 

7. 

8. 

9. 

10. 

-7 

+7 

+7 

-7 

12 

~5 

+5 

-5 

+5 

-9 

11. 

12. 

13. 

u. 

15. 

—a 

-la 

-9a 

-12  a; 

+  12  a; 

-6a 

4-3  a 

-4  a 

-Qx 

+6  a: 

16. 

17. 

18. 

19. 

20. 

—  (J?X 

-dx'^y 

—  12mn 

-\-ab 

+  17^:^ 

7  A 

-\-12xhj 

—9mn 

-  20  ab 

23. 

-3a:^ 

21 

22. 

-%x- 

-22/) 

-\-^{x'-f) 

l{x^Jr2xy-\-f) 

+4(x- 

-22/) 

-\-9{x'-y') 

-A(x^+2xy-\-y^) 

ADVANCED    ARITHMETIC.  343 

Copy  and  subtract  : 

24.  From3tt-562-c  take  7  a+4 62-10 c. 

25.  From  7 a^-\-9xy-Qb^  take  nb^-\-4.xy-10a\ 

26.  From  —75{x^y—xy^)  take  —160{x^y—xy^). 

27.  17a:-52/+6-(5a;+ll2/-10)=(   )? 

28.  9xy-{7ax-^12xy-Sz)  =  {   )? 

29.  7x^y-Sxy'^-{-y^-{-7i/+Qxy''-9yx^)=(   )? 

30.  a-\-b  —  c  —  d—x  —  {a  —  b-\-c—d—x)  =  {   )? 

31.  x^-j-2xy-{-y^-{x^-2xy^y^)  =  (   )? 

^^.  x^+Sxhj-i-Bxy^-\-y^-{x^-Bx^y+Bxy^-f)  =  (   )? 

163.  Mviltii^lication. — When  the  multiplier  is  posi- 
tive, it  shows  how  many  times  the  multiplicand  is  to  be  taken 
as  an  addend  (see  p.  21) ;  but,  when  negative,  it  shows  how 
many  times  the  multiplicand  is  to  be  taken  as  a  subtrahend. 

EXAMFIiES. 

1.  Multiply  +7  by  +3. 

Note. — Starting  with  0,  we  add  +7  three  times. 

Process  :  (1)  0+(  +  7)=  +7,  used  once. 

(2)  +7-t-(+7)=  +14,  used  twice.  , 

(3)  + 14 + ( + 7)  =  +  21 ,  used  three  times . 
.'.   +3x+7=+21,  result. 

2.  Multiply  -7  by  +3. 

Note. —  Starting  with  0,  add  -7  three  times. 
Process:  (1)  0+(-7)=  -7,  used  once. 

(2)  -7+(-7)=  -14,  used  twice. 

(3)  - 14  -|-(  -  7)  =  -  21 ,  used  three  times. 
.*.  +3x  -7=  -21,  result. 

3.  Multiply  +7  by  -3. 

Note.— Starting  with  0,  subtract  +7  three  times. 
Process  :  (1)  0-(  +  7)=  -7,  used  once. 

(2)  -7 -(  +  7)= -14,  used  twice. 

(3)  - 14  -  (  +  7)  =  -  21 ,  used  three  times. 
.-.   -3x+7=  -21,  result. 


344  ADVANCED    ARITHMETIC. 

U.  Multiply  -7  by  -8. 

Note. —  Starting  with  0,  subtract  -7  three  times. 

Process  :  (1)  0-(-7)=  +7,  used  once. 

(2)  +7 -(-7)= +14,  used  twice. 

(3)  +14-(-7)= +21,  used  three  times. 
.-.    _3x-7=+21. 

Let  us  compare  these  examples: 

(1)  +3x+7=+21. 

(2)  +3x-7=-21. 

(3)  -3x+7=-21. 

(4)  -3x  -7= +21. 

Observe  (1)  that  the  numerical  products  are  the  same  (each  21,  or  the 
product  of  3  and  7),  and  (2)  when  the  signs  of  the  factors  are  alike  (both 
plus  as  in  (l)or  both  minus  as  in  (4) ),  the  product  is  positive;  but  when  the 
signs  are  unlike  (as  in  (2)  and  (3)  ),  the  product  is  negative. 

Law  of  Signs  in  Multiplication  :  Like  signs  give  plus ;  un- 
like signs  give  minus. 


5,                          6. 

7. 

8. 

+  12  «                   +12  a 

-12a 

-12a 

+66                     -66 

+6  6 

-66 

+72  a6,  result.     — 72  a6,  result. 

—  72  a6,  result. 

+72  a6,  result. 

9.   -7aX+56x-8c=(   )? 

Process:   (1)   -7 ax +5  6  = 

-35a6. 

(2)   -35a6x-3c 

=  +105  a 6c,  result. 

10.  Multiply  4  m— 7  n  by  5  71. 

Process  :  4m  -In 
-5n 

-20mn+35w2;  or, 
S5 n^ -20 mn,  result. 

Note. — In  writing  polynomials,  it  is  customary  (but  not  necessary)  to 
write  a  positive  term  (if  there  be  one)  on  the  left,  and  omit  its  sign. 


ADVANCED    ARITHMETIC.  345 

11.  Multiply  ba(x-\-y)  by  —h. 

Note. — This  multiplication  may  be  performed  (1)  by  multiplying  the  5  a 
by  -h,  (2)  by  multiplying  {x  +  y)  by  -h,  or  (3)  by  removing  the  paren- 
theses from  the  multiplicand  and  multiplying  each  term  by  -  h.     Thus : 

(1)  (2) 

ba{x+y)  5a  {x+y) 

-  b  -b 


-5ab  {x+y),  result.  6a  {-bx-by),  result. 

(3)  ba  {x+y)=bax+5ay 

-b_ 

-babx-haby,  result. 

12.   {a-b)^={a-b){a-b)  =  (   )? 

Process  :   a-b 
a-b 
-ab+b^ 
a^-ab 


a^-2ab  +  b^,  result. 

Principle:  1.  The  square  of  the  difference  of  two  numbers  is 
equal  to  the  square  of  the  first  minus  twice  the  product  of  the  first 
times  the  second  plus  the  square  of  the  second.     (Commit.) 

Note. — Compare  this  principle  with  the  one  on  page  181,  noting  the 
points  of  similarity  and  difference. 

13.  Multiply  a +6  by  a  — 6. 

Process :   a  +  b 
a-b 

+ab-b^ 
a^  —  ab 


a^        -  6^,  result. 

Principle  :  2.  The  product  of  the  sum  and  difference  of  two 
numbers  is  equal  to  the  square  of  the  first  minus  the  square  of  the  sec- 
ond.    (Commit.) 


346  ADVANCED    ARITHMETIC. 


Uf..  Multiply  3  a— 2a:  by  5  6— a. 

Process:  3a -2  x 

56    —a 


15ab-10bx-Sa^+2 ax,  result. 


EXERCISE  OXLI. 

1.  What  change   in   the   definition  (p.  21)    is   necessary  to 
make  it  include  the  multiplication  of  negative  numbers  ? 

2.  What  does  a  positive  multiplier  mean  ? 

3.  What  does  a  negative  multiplier  mean  ?     Give  the  law  of 
signs. 

If.  Repeat  principle  1.     In  what  respect  does  it  differ  from 
that  on  p.  181  ? 

5.  Repeat  principle  2. 

Multiply  (orally)  : 

6.  -5X+12.  10.  -aX-\-b.  U.  (a+6)2=(   )? 

7.  +6X-10.  11.  -2aX-3  6.  15.  {a-by  =  (    )? 

8.  -7x-9.  12.  -i-5aX-\-7m\  16.  \x-yY={    )? 

9.  -8X+11.  13.  -4:X^Xl2y'^.  17.  {x-ij)(x-\-y)  =  {   )? 

Copy  and  multiply : 

18.  —9  (x—ij)  by  —5  a. 

19.  7x-12y  by  a  —  b. 

20.  24m2  — 7mnby  6  m— lln. 

21.  ^Q>xy—bxy'^hy%x—bz. 

22.  x^-\-2  xy-\-y'^  by  x-\-y. 

23.  x^  —  ^  x^y-\-^  xy^—'f  by  x^—y^. 
21f.  ^xy  —  7am-^n^hj6a  —  7b. 

164.  Division. — By  principle  3  in  division  (page  25), 
the  dividend  equals  the  product  of  the  divisor  and  quotient. 


ADVANCED    ARITHMETIC.  347 


Since  (1)  +7x+3=+21,  +21 
Since  (2)  -7x+3  =  -21,  -21 
Since  (3)  -7x-3  =  +21,  +21 
Since  (4)  +7x-3=-21,  -21 


+7  =  +  3. 
-7== +3. 
-7=^3. 

+7=-3. 


Observe  that  when  the  signs  of  the  dividend  and  divisor  are  alike  the 
quotient  is  positive,  and  when  the  signs  of  the  dividend  and  divisor  are 
unlike  the  quotient  is  negative. 

Law  of  Signb  in  Division  :  Like  signs  give  plus ;  .unlike  signs 
give  minus. 

EXAMPIiES. 

1.       ,  2.  3.  U- 

6)24  -6) -24  -6)24  6) -24 

4,  result.  4,  result.        —4,  result.  —4,  result. 

5.  6.  7.  8. 

4y)24xy        —Qx)—24xy_        -\-Qy)—24:xy         —4y)24xy 


6x,  result. 

42/, 

result. 

—4  a:,  result. 

-6  a;,  re 

9. 

10. 

Qy)12xy-Q 

2x  -y 

,  result. 

—^xil^ax  —  lOxy 

— 3a+22/;  or, 

11.  Divide  a:2- 

-2xy-\-y 

2  by  x—y. 

2y-Sa, 

result. 

Process  : 

x-y)  x^- 

2xy  +  y%x- 
-xy 

-xy+yl 
-xy+y^ 

y,  result. 

Explanation  :  (1)  Find  how  many  times  the  first  term  of  the  dividend 
contains  the  first  term  of  the  divisor.  x^  +  x  =  x.  Place  x  in  the  quotient. 
Multiply  the  whole  divisor  by  the  quotient  x,  and  subtract  the  product 
from  the  dividend. 

(2)  Proceed  just  as  before.  Find  how  many  times  the  first  term  of  the 
remaining  dividend  contains  the  first  term  of  the  divisor.  -xy-*-x=  -y. 
Place  -yin  the  quotient.  Multiply  the  whole  divisor  by  -y&nd  sub- 
tract.    Result,  a: -2/. 


348  ADVANCED    ARITHMETIC. 

12.  Divide  a;2  — 2/2  by  a;+2/- 

Process  :   x+y)  x^-y^  i^-y-i  result. 
x^-\-xy 


Note. — When  we  go  to  subtract  +xy,  there  is  no  similar  term  in  the 
dividend ;  so  the  remainder  is  -  xy,  to  which  we  bring  down  the  next 


EXERCISE  CXLII. 

1.  What  is  the  laiv  of  signs  in  division  ?     Compare  with  the 
law  of  signs  in  multiplication. 

Divide  (orally) : 

2.  25^5.  7.   —26a^x-^—6ax. 

3.  -25^5.      •  8.   -36a;22/2^9a;2/. 
^.  25-^—5.  9.  17 ax^y^axy. 

6.   —25-^—5.  10.   — 12  ax(7n-\-n)  ^12  ax. 

6.    -10ah^-\-ah.  11.   -12ax  (y-z)^(y-z). 

Copy  and  divide  — 

12.  bw?n— 4:7)171^  by  mn. 

13.  lhx'^-2bxy^hybx. 
U.  x^-\-2xy-\-y^  hj  x-\-y. 

15.  x^—2ax-\-a^  hj  x—a. 

16.  x^-^x'^a^^xa^-a^hj  x^-2ax^a^. 

17.  x^-\-^x'^y-\-%xy'^-\-y^hj  x-{-y. 

1 65 .   Factoring. —  The  factors  of  numbers  may  be  found 
by  inspection  or  by  division.     (See  pp.  40  and  41.) 

EXAMPLES. 

1.  Factor  bnh. 

Process  :  5  a&  =  5  x  a  x  6. 

Factors,  5,  a,  and  b. 


ADVANCED    ARITHMETIC.  349 


2.  Factor  25a '-^c  (Inspection). 

Factors,  5,  5,  a,  a,  and  c. 

3.  Factor  bh^^{x-\-y). 

Factors,  5,  6,  h,  and  x-\-y. 

^.  Factor  5 a6- 10 62. 

Process :  5  h  )  5  a6  - 10  6^ 
a  -  26. 
Factors,  5,  h,  and  a-2h. 

5.  Factor  a'^-\-2ah-\-h'^ . 

Note. — By  the  principle  (p.  181),  a^+2ah  +  h'^  is  the  square  of  the  sum  of 
two  numbers. 

Process:  a^+2ah  +  h^  =  {a  +  h)  {a  +  h). 
Factors,  a+h  and  a +  6. 


6.  Factor  a2-2a.6+62. 

Note. —  Apply  prin.  1,  p.  345. 

Factors,  a-h  and  a-h. 


7.  Factor  a^ - 

62. 

Note.— Apply  prin.  2,  p.  345. 

Factors,  a  +  6  and  a-h. 

EXERCISE  CXLIII. 

Factor  — 

1.   12  a. 

5.  ^^xHf. 

9.   12(x-y). 

2.  1562. 

6.  51  mnl 

10.  x(7n-\-7i). 

3.  2b  ab. 

7.  64^3^2. 

11.   16:c7/(a4-6). 

If..  12mP'n. 

8.   121  x^y^z. 

12.  75a;2(a+6)2. 

13.  m2+2 

'  71171 -\-n'^. 

16.  x^ 

-2a:+l. 

i^.  a;2-2 

xz-\-z\ 

17.  2;2 

-2/2. 

15.  a^^2i 

x+1. 

18.  x^- 

-%x\j^'^xf-y^. 

350  ADVANCED    ARITHMETIC. 

166.   Greatest  Common  Divisor. — The  G.  C.  D  is 

usually  found  by  factoring.     The  plan  here  is  not  different 
from  that  found  on  page  47. 

EXAMPLES. 

1.  Find  the  G.  C.  D.  of  5ac,  2b  ax,  lb  an. 

Process  :  (1)  5ac  =  5xaxc. 

(2)  25ax  =  5x5xaxx. 

(3)  15an  =  3x5xaxn. 
.-.  G.  C.  D.  =  5xa  =  5a. 

2.  Find  the  G.  C.  D.  of  ba{x-y),  Ibmx-Wmy. 

Process:  (1)  5a{x-y)  =  5xax{x-y). 

(2)  l5mx-15iny  =  Sxbxmx{x-y). 

.•.   G.G.I>.  =  5x{x-y)  =  5{x-ij);  or, bx-5y. 

3.  Find  the  G.  C.  D.  of  x'^+^xy+y^,  x^-y^,  7x-\-7y. 

Process:  (1)  x^+2xy  +  y^  =  {x-\-y)  {x+y). 

(2)  x^-y^  =  {x+y){x-y). 

(3)  7x+7y  =  7{x+y). 
.'.  G.  G.  'D.=x+y. 


> 


EXERCISE  OXLIV. 

Find  the  G.  C.  D.  of— 

1.  a2,  5ax,  Say.  3.  24x^y^,  Wxy^,  40x^yz^. 

2.  7mx,  14am,  Sbmy.         4-   12a'^m'^n^,  IS am^n,S6a^m^n^. 

5.  Sax(m—n)j  m^—n^,  7m  — In. 

6.  2bax{y-z),  Ibay^  —  lbaz^,  lOay-lOaz. 

7.  x^-\-2xy-{-y^,  x'^-ij^,  17ax-\-17ay. 

8.  (x-y)^,  x^—y^,  bhx—bby. 

167.  Least  Com^mon  Multiple.— The  L.  C.  M.  is 

usually  found  by  factoring.     The  plan  here  is  not  different 
from  that  found  on  page  49. 


ADVANCED   ARITHMETIC.  351 

EXAMPXiES. 

1.  Find  the  L.  C.  M.  of  5  a,  10 ay,  l^a^y'K 

Process:  (1)  5a  =  5xa. 

(2)  10a^  =  2x5xax?/. 

(3)  15aV  =  3x5xaxaxi/Xi/. 

.'.   L.  0.  M.  =  2x3x5xaxax?/x?/  =  30aV- 

2.  Find  the  L.  C.  M.  oi6a{x-y),  7ab,  ^^a(x-y)^. 

Process:  (1)  5a{x-y)  =  5yax{x-y). 

(2)  7ab  =  7xaxb. 

(3)  35a{x-yf  =  5x7xax{x-y){x-y). 

.-.  L.  CM.  =5x7 X axbx{x -y){x -y)  =  36 ab{x - yY ; or f 
3ba{x^-2xy-\-y^). 

EXERCISE  CXLV. 
Find  the  L.  C.  M.  of— 

1.  7a,  Uah,  2S a^b. 

2.  9?/2,  18x2/,  24a;22/2. 

S.  2mn^z,  Qm^nz,  12mnz^. 
4-'  ms^,  5  mns,  20  xns. 

5.  \.bax(m—n),  25,  10a{m—n),  6x{m^—n^), 

6.  12,  X2-2/S  {x-yy,Q{x-y). 

7.  86,  12c,  862c,  4:b{m-n). 

8.  x^  —  2xy-{-y^,x^-\-2xy-\-y'^,x^—y^. 

B.     FRACTIONAL    NUMBERS. 

168.   Reduction. — The  reduction  of  fractions  here  does 
not  differ  in  principle  from  that  found  in  Articles  35  to  39. 

EXAMPLES. 

5  a 
1.  Reduce  ^^  to  a  fraction  whose  denominator  is  10  6a;. 

Z  0 

Process:   (1)  10 5a: -5-2 6  =  5. r. 

--,.  bxxba     25  ax  ,^ 


352  ADVANCED    ARITHMETIC. 

4Sax'^y 

2.  Reduce    -^    „    , — - — r-  to  its  lowest  terms. 

Note.— 24a:c  is  the  G.  0.  D.  of  the  numerator  and  the  denominator. 

Process:     -^    »  .   ^ — ,-—^. —  =  _    ,    ^ — r,  result. 
72  a^x{m -n)-^24ax    3a{m- n) 

3.  Reduce  t>x—a to  the  form  of  a  fraction. 

xy 

r,  r.  ab     xy  (5x-a)-ab    6 x^y - axy -ah 

Process:   ox -a =  — = — ^ ^ .result. 

xy  xy  xy 

Note.— (1)  Multiply  the  integral  part  by  the  denominator,  (2)  add  the 
numerator,  and  (3)  place  the  sum  over  the  denominator. 

A.  Reduce  -, to  a  mixed  number. 

a  —  b 

252 

Process:   a-b)a^-2ah-\-Sb^{a-b-\ 7,  result. 

gg  -ab  «-^ 

-a6+3  52 
-ab    +62 


+2  6^,  remainder. 

5.  —  ,77-,  — ^"1-  to  least  common  denominator. 
a    oc    a^d 

Note. — L.  0.  D.  of  a,  6  c,  and  aH  is  Qahd. 

Process  :   (1)  6  a^ca -^  a  =  6  aca;  - — r— »    o    .« 

^  '  6  acd  Xa     6  a^cd 

io\  a    2  ^    a         2^    a^dx2b_2a^d 

iS)  Qa^cd^aH  =  6c;p^  =  ^^^. 
'  ^cxa^d     Qa\d 

EXEKCISE  CXLVI. 

Reduce — 

1.  5  to  a  fraction  whose  denominator  is  x. 

2.  —  to  a  fraction  whose  denominator  is  7  amz. 

VIZ 


ADVANCED   ARITHMETIC.  353 

3.  r^   n     -i  to  lowest  terms. 
50  a^my"^ 

12xy(m—n)     ,     ,  ,  , 

^-    iQ   2/    o iT  to  lowest  terms. 

5.  ^  \  ^y^y    ^o  lowest  terms. 

/^2 5'»2 

^.  ; — —  to  mixed  number. 

x-\-y 

^    x^-\-2xy-2y^  ^        .      ,  , 

7.  — ' r ^^  to  mixed  number. 

x+y 

8.  a  2 to  fractional  form. 

c 

9.  b^v—m-\ — —  to  fractional  form. 

^  x^y 

10.  Ti—1-L  to  L.  C.  D. 
0   ac    ah 

n.  7,  -,  ^,  I  to  L.  C.  D. 

x    bo 

,^     6x     7b           7z         J.     T    r^   T^ 
12.  -^,  -^,  —7 r  to  L.  C.  D. 

a^y  ay^   ay(m—n) 

169.   Addition, — Addition   of  fractions  here  does  not 
differ  in  principle  from  that  found  in  Article  40. 

EXASIFIiES. 

^     *Ti3a:  5x  7 X 

„              S.X    5x    7x     lSx+15x+Ux     47  ,^ 

Process  :  -4- +"8  +  ^2  = 24 "  24^'  result. 

Note  :    —x  is  the  same  as-^rj-. 
24  24 

„  a      c   ,    c      5a/^+6c+ac  ,^ 

Process:  -r-\ — r+^T  = ^-^ ,  result. 

b     ab    5b  5ab         ' 


354  ADVANCED    ARITHMETIC. 


S,  Add  26,  H ,  — 


X         m—n 

Process:  2h-\ = 

X     m-n 


2bmx  —  2bnx     7  am -7  an      Sbmx 
xm  —  xn  xm-xn       xm-xn 

2bmx-2bnx-{-7  am-7  an-Sbmx 

xm  —  xn  ~ 

7  am  -  2  bnx  -7an-Q  bmx 


xm-xn 
Note. — Have  the  pupil  explain  each  step. 


,  result. 


EXERCISE  OXLVII. 
Add: 

^'  6'  ~^4'  "^  8*  ^'  "5"'  6~'  "^TO"' 

0  ?^    __^    4-L  ff    5a;-52/        7a;-3y 

^-     a;'       2/'      a:2/'  ^-        8       '  12     * 

^  a;-2/   a;+2/  ^    S^I^;  ^        3a;-6 


7. 


2    '     2    •  a   '    '  d 

5  3  x—y 


Sx+Sy'       5a:+52/'  (a^+2/)^* 


^    5a     7a6_8a^ 
mc     om       4  c 

170.    Subtraction. —  Subtraction  of  fractions  here  does 
not  differ  in  principle  from  that  found  in  Article  41. 


EXAMPLES. 


3*  3/ 

2.  From  -r  take  -=. 
4  5 


^  X      X    5x     4:X     X  ,^ 

iVocm.—  -g=-- 20  =  20- '•"suit. 


ADVANCED    ARITHMETIC.  855 


2.  Prom  -Y  take  —  -~. 


Process  :(1)  'J -[-*-l]=^-^+*l.    (Why?) 


a 


bx+Sy]_ 


6a 


j=( )? 


a       [      6a     J      6a       [      6a     J 

^ —  =  — ^ — -,  result. 

6a  6a 

6a:— 5,    ,     6a:— 5 
4.  From take  — ; — . 

x-y  x-\-y 

p  _  6.r-5    Qa:-5_6x^-bx+Qxy-6y    6ae^--5.r-6y  +  5y_ 

X-y      x+y  x^-y^  x"-y^ 

^x^-hx-\-Qxy-hy-{Qx^-bx-Qxy-\-by)_ 

x^  —  y^ 

Qx^-hx-^Qxy-hy-Qx^-^bx^-Qxy-hy  _l2xy-lQy 

o      o o      o —  J  rGSUlt. 

^2  _y2  x^  —  y^ 


EXERCISE  GXLVIII. 


1.  From  -=^take-7r-. 

5  9 

^    ^  12y  ^  ,      51/ 

2.  From   — ^  take  ^^^ 

a;  11a: 

7  10 

5.  From   -take 


y         132/* 

a*  a*'^ 

4-.  From  —  take^r"- 

2/  72/ 

^    „  a4-^  ^  1     CL  —  b 

5.  From  -tt^  take  — r^-, 

12  lo 

6,  From  — ; — take . 

x-\-y  x  —  y 


856 


ADVANCED   ARITHMETIC. 


7.  From 


12x 


10,  ,       6a;-5 
take 


8.  a  — 


x-\-2y  Sx-\-Qy 

36       {7  a- 


10      I 


^-h^  = 


=  (    )? 


171.   Multiplication. — Multiplication  of  fractions 
here  does  not  differ  in  principle  from  that  found  in  Article  42. 


i:XAMPIiES. 


1.  Multiply^  by  1^. 

Process :  —  x  -^  =  h~»  result. 
ax      Q      2  .r' 

Note.— Cancel,  when  possible. 
a  —  b     m-\-n 


a-\-b      a  —  b 
Process  : 


={  )? 


— -r  X =  -—J- ,  result. 

a+b     a-b      a+b 


3.    "^X 

b 


^-X  —  =  i    )? 


„  a  ^     c  ^  a 

Process  :   t-  x  -  t-  x  i;- 

b         b     Sc 


362 


,  result. 


EXERCISE  CXLIX. 


Multiply — 

''   b  ^^  Wc 

2.  ^  by  P-. 
32/        12  ly 

5a6  ,      l^xz 

^'  Uy  ^  10^' 


i.   Sab  hy 


5m 
7  ax' 


Saxy         llb'^y 

22  bh  ^  12  ax^' 

^    x-\-y  ,     x—y 
6.   —r^  by ^. 

x-j-z    "^  x—z 

x{a-^b)  a-b  _ 

''       a-b        ^  12(a;+a)- 

5a— 66        5m— 3n 
^'   3.T+6  7/^  5?i-36' 


ADVANCED   ARITHMETIC.  357 

172.    Division. — Division  of  fractions  here  does  not  dif- 
fer in  principle  from  that  found  in  Article  43. 

EXAMFIiES. 

1.  Divide  —  by  — ^. 

n  5  a    Sb    6  a     m^     bam  ,, 

Process  :  —  -^  — « =  —  x  irr  =  -ttti  result. 
m      m'     m      Sh       3  6 

2.  Divide  — I —  by  — ; — . 

m-\-n    ''    x-\-y 

_,  m-n    m-n    m-n     x+y      x-\-y  .^ 

Process  :  — --  +  — ; —  =  — --  x —  =  — -^,  result. 

m+n     x+y     m+n    m-n    m+n 


3.  Divide  -. —  by  56. 
4  m    ^ 


Process  :  -. —  +  56=  -. —  x  -—  =  ,  result. 

4  m  4971     56    206m 


EXERCISE  CL. 


Divide — 


1x,     92  ^    82-4,     42-2 

-5-^^ To-  ^-  -y^^^^- 

252     •^3522-  m+?i    •^2m+2ri 

1422  ^     52a;  12-9x 


21  ay     -"  ^my  4       ^ 

5  a6c        20 a6c2  '   16-12 a;     ^    ^ 

7xyz    ^  12a;2m  x-\-y 


II.    STUDY  OF  PROBLEMS. 
A.    PROBLEMS  OF  ONE  BASIS. 

173.  Solving  Equations  of  One  Unknown 
Number. —  We  have  learned  that  various  operations  may  be 
performed  on  an  equation  without  destroying  the  equality : 

(1)    The  terms  may  he  transposed.     (See  p.  81.) 


358  ADVANCED    ARITHMETIC. 


EXAMPLES. 


l.ox—lb=4:X—b.     Find  the  value  of  ic. 

Process  :  (1)  5  .r  - 15  =  4  c  -  5. 
Transposing,  (2)  5x-4x  =  16-b. 
(2)  =  (3)  .T  =  10,  result. 


(2)  An  equation  may  he  multiplied.     (  See  p.  83.) 

2.   - 
4 


X        X 

-  =7.     Find  the  value  of  x. 


Process:  (1)  — =  7. 

4      5 

(1)  =  (2)  S^zif  =  7. 

20x(2)  =  (3)  5x-4.r  =  140. 
(3)  =  (4)     a:  =  140,  result. 

Principles  :  1.  Any  equation  may  he  cleared  of  fractions  hy  he- 
ing  multiplied  hy  the  L.  C.  M.  of  the  denominators  of  all  the  frac- 
tions. 

S,  1 'L^ni=is.     Find  the  value  of  x. 

8         7 

Process  :  (1)  —  -  ^^^-^  =  13. 
8  7 

56x(l)  =  (2)  49 ^-(48 .1+32)  =  727. 

(2)  =  (3)  49 .1-48.*: -32  =  727. 

Transposing,  (4)  49.r-48.r  =  727  +  32. 

Collecting,  (5)  .r  =  759,  result. 

2.  The  signs  of  all  the  terms  of  an  equation  may  he  changed  hy 
multiplying  the  equation  hy  —1. 

4.  5.c+3=6.c-9.     Find  the  value  of  a;. 
Process:  (1)  5a:+3  =  6a;-9. 
Trans.,  (2)  5a:-6:r= -9-3. 

Col.,  (4)   -.r=-12. 
-lx(4)-(5)  .r  =  12,  result.        , 

Note. —  In  practice,  we  simply  rewrite  the  equation  with  the  signs 
changed. 


ADVANCED   ARITHMETIC.  359 

(3)  An  equation  may  be  divided.     (See  p.  89.) 
5.  35a;-15=18x+70.     Find  the  value  of  x. 

Process:   (1)  35a;- 15  =  18 a;+70. 
Trans.,  (2)  35a:- 18:?:  =  70 +15. 

Col.,  (3)  17 a:  =  85. 
(3) +  17  =  (4)  a;  =  5,  result. 

Plan:  In  finding  the  value  of  the  unknown  number  in  an 
equation,  it  is  best  to  follow  a  definite  plan.  The  operations 
required  should  usually  be  performed  in  the  following  order : 

(1)  Clear  of  fractions. 

(2)  Clear  of  parentheses. 

(8)  Transpose  all  terms  containing  the  unknown  number  to 
the  left  member  of  the  equation,  and  all  other  terms  to  the 
right  member. 

(4)  Collect  the  terms  of  each  member  into  one  term. 

(5)  Change  the  signs.* 

(6)  Divide  by  the  coefficient  of  the  unknown  number. 

Note. —  It  is  not  often  that  all  these  operations  will  be  required  in 
solving  one  equation,  but  it  is  usually  best  to  perform  those  that  are  re- 
quired in  the  order  indicated. 

EXERCISE  OLI. 

1.  What  is  an  equation  f 

2.  What  principles  are  involved  in  the  transposition  of  the 
terms  of  an  equation  (page  81)?  Give  the  law  of  transposition 
(page  82).         . 

3.  What  is  the  rule  for  removing  parentheses  (page  341)  ? 
4-  Give  principle  for  clearing  of  fractions. 

5.  Upon  what  principle  may  the  signs  of  all  the  terms  of  an 
equation  be  changed  ? 


'The  (5)  may  be  dispensed  with,  by  dlTiding  by  the  negative  coefficient  in  the  (6). 


860  ADVANCED   ARITHMETIC. 

Find  the  value  of  the  unknown  number  in — 

7.  25-5a;=0.  ^'    5  "^  6  "^  3         * 

8.  72/+12-482/.  ^_    5^_§^-iii 
^.  5a;-18-ja;-38=6.  6       11  ~ 

11.  5 71+ 164-871= 10  n.  7       4  ^ 

^g.  _^+i,=51-68.  7^+80-^-^-205 

i5.  24a-12i-6ia=5i.  ^-'-   '  ^+*^"-  4  -'^^»- 

-1-1=8-  .,.8i^-%^=_74. 

m    m_  5  m+5    6m-7_9  7?i-5i 

^^'    4+8  -'^^-  ^^-  ~4  3~  -^6~" 

174.  Problems. 

EXAMPLES. 

i.  J  of  a  number,  \  of  the  number,  and  10,  are  together 
equal  to  the  number.     Find  the  number. 
Solution :  Let  .r=  the  number.* 
Then,  (1)  lx  +  lx+10  =  x.    (Why?) 
12x(l)  =  (2)  4a;+3a;+120  =  12a;. 
(2)  =  (3)  12a;=120+4a;+3a;. 
Trans.,  (4)  12x-4x-Sx  =  120. 

Ool.,  (5)  5x  =  120. 
^of(5)  =  (6)     «  =  24,  answer. 

2.  Find  a  number  whose  J  exceeds  its  i  by  150. 
Solution  :  Let  a:  =  the  number. 
Then,  (1)  ^.r-i.r  =  150.     (Why?) 
12x(l)  =  (2)  4;r-3.r  =  1800. 
Col.,  (3)     a:=  1800,  answer. 

*  "Let  x=the  number,"  means  that  x  Is  to  be  used  to  represent  the  number.    The  sign  =  is 
used  because  it  is  more  convenient  than  the  word  represent. 


ADVANCED    ARITHMETIC.  861 

3.  The  sum  of  two  numbers  is  25,  and  their  difference  is  5. 
Find  the  numbers. 

Solution :  Let  x=  the  smaller  number. 
Then,  25  -  a;  =  the  larger  number 
and  (1)  25-a;-a:  =  5.     (Why?) 
Trans.,  (2)  -a;-a;  =  5-25. 
Ool.,  (3)  -2a;=-20. 
(3)^-  -2  =  (4)  .T  =  10,  smaller  number. 

(5)  25- a;  =  15,  larger  number. 

j^.  What  four  successive  integers  are  together  equal  to  46  ? 

Solution :  Let  x=  1st  number. 
Then,  a:  +  l  =  2d  number, 
:r+2  =  3d  number, 
a;+3  =  4th  number, 
and  (1)  :i:+(:r+l)+0r+2)  +  CT+3)  =  46.     (Why?) 
(1)  =  (2)  x+x+l+x^-2+x+^  =  4Q. 
Trans.,  (3)  a:+a;+a;+:r  =  46-l-2-3. 

Col.,  (4)  4 a:  =  40. 
\  of  (4)  =  (5)  a;  =  10,  Ist  number. 

Then,  the  other  numbers  are  11,  12,  and  13, 

5.  Henry  spent  yV  of  his  money,  and  then  received  $65 ;  he 
then  lost  -f  of  all  his  money,  and  had  in  hand  $10  less  than  at 
first.  •  How  much  had  he  at  first  ? 

Solution  :  Let  a;  =  his  money  at  first. 
Then,  ^^x  =  v,TnX,.  left, 

T\a;+65  =  amt.  after  receiving  $65, 
and  (1)  i(3»i^+65)  =  :r-10.     (Why?) 
(1)  =  (2)  iiX+^  =  x-lO. 
44X(2)  =  (3)  9:c  +  715  =  44^-440. 
Trans.,  (4)  ^x-Ux=  -440-715. 
Col.,  (5)   -35a:  =-1155. 
(5)-t--35  =  (6)  :k  =  33. 

Answer,  $33. 

6.  Goods  sold  at  10%  gain ;  if  the  goods  had  cost  $120 
more,  the  loss  would  have  been  10%.     Find  the  cost  price. 

Note.— The  selling  price  is  110^  or  {^  of  the  real  cost,  and  W  or  ^^y  of 

the  supposed  cost. 


ADVANCED   ARITHMETIC. 

Solution :  Let  x  =  real  cost. 

Then,  .r+ 120  =  supposed  cost 
and  (1)  H:t'  =  xU^+120).    (Why?) 
10x(l)  =  (2)  ll.r  =  9a:+1080. 
Trans.,  (3)  11  a: - 9 a;  =  1080. 

Col.,  (4)  2 a:  =1080. 
^of  (4)  =  (5)  a;  =  540. 

Answer,  $540. 

7.  By  selling  my  watch  for  $36,  I  lose  |  of  its  cost.  What 
did  it  cost  ? 

Equation:  a; -fa:  =  36.     (Why?) 

8.  A  and  B  have  equal  sums  of  money ;  A  gains  $100  and  B 
loses  $150;  then,  twice  A's  money  is  equal  to  three  times  B's. 
What  sum  had  each  at  first  ? 

Equation:  2  (a: +100)  =  3  (a: -150).     (Why?) 

P.  In  a  certain  weight  of  gunpowder  the  saltpetre  is  6  lb. 
more  than  -J  of  the  whole,  the  sulphur  is  5  lb.  less  than  i  of 
the  whole,  and  the  charcoal  is  3  lb.  less  than  i  of  the  whole. 
Find  the  entire  weight  and  the  weight  of  each  part. 

^g'wah'on :  Let  a:  =  whole  weight. 
Then,  (1)  a-x+Q)  +  {lx-b)  +  {ix-S)  =  x.    (Why?) 

10.  6  times  the  ratio  of  ^  to  j  is  equal  to  how  many  times  the 
ratio  of  3  to  9  ? 

Solution  :  Let  a:  =  required  number. 

Then,  (1)  6x(^:|)  =  a;x(3:9).     (Why?) 

(1)  =  (2)  6xf  =  .rx^. 

(2)  =  (3)ir=|. 

3  X  (3)  =  (4)  a:  =  8,  answer. 

11.  A  and  B  can  perform  a  piece  of  work  in  10  days.  They 
both  work  three  days  and  B  then  finishes  it  in  12  days.  How 
long  will  it  take  each  to  do  the  work  ? 


ADVANCED   ARITHMETIC.  863 

Equation  :  Let  :c  =  time  reqd.  by  A  to  do  the  work. 
Then,  —  =  part  done  by  A  in  1  da. , 

iV  =  part  done  by  both  in  1  da. , 
f%  =  part  done  by  both  in  3  da. , 

-^  —  =  part  done  by  B  in  1  da., 

12     12 

— =  part  done  by  B  in  12  da. , 

10      a:     ^ 

and(l)^+i?-i?  =  l.    (Why?) 

Note.— The  question  is  sometimes  asked,  *' Where  did  you  get  the  i.^'* 
The  concrete  form  of  the  equation  will  explain : 

3         12         12 

—  of  work +—  of  work of  work  =  1  x  the  work. 

10  10  X 

12.  If  I  subtract  $20  from  ^  of  my  money,  multiply  the  dif- 
ference by  7,  and  subtract  this  product  from  $1800,  the  result 
will  be  equal  to  my  money.     How  much  have  I  ? 
Equation:  1300-7(ix-20)  =  :c.     (Why?) 

Note. —  Always  require  the  pupil  to  explain  his  reasons  for  forming 
the  equation ;  because  there  is  where  the  pupil  does  his  investigative 
thinking.     Solving  the  equation  is  mechanical. 


EXERCISE  CLII. 

i.  "i,  ^  and  i  of  a  number  exceed  the  number  by  14:  find  the 
number. 

2.  A  number  decreased  by  its  \  and  f  equals  54 :    find  the 
number. 

3.  A  number  increased  by  its  ^  and  -^  equals  171 :   find  the 
number. 

J,..  I  of  a  number  is  80  greater  than  its  ^'.  find  the  number. 

5.  5  times  a  certain  number  is  27  more  than  its  i :  find  the 
number. 

6.  The  sum  of  8  successive  numbers  is  72:  find  the  numbers. 


364  ADVANCED   ARITHMETIC. 

7.  The  sum  of  5  numbers  each  differing  from  the  next 
smaller  by  4  is  140:  find  the  numbers. 

8.  The  sum  of  3  numbers  each  3  times  the  next  smaller  is 
39:  find  the  numbers. 

9.  A  and  B  had  the  same  amount  of  money ;  A  lost  $40  and 
.  B  gained  $60;  i  of  A's  money  was  then  equal  to  |  of  B's  :  what 

did  each  have  at  the  start  ? 

10.  A  and  B  had  the  same  amount  of  money ;  A  gave  B  $10; 
B  then  had  5  times  as  much  as  A:  what  sum  had  each  at 
first  ? 

as  much  as  A :  what  sum  had  each  at  first  ? 

11.  In  the  composition  of  a  quantity  of  gunpowder,  the  nitre 
was  10  lb.  more  than  f  of  the  whole,  the  sulphur  was  4^  lb.  less 
than  J  of  the  whole,  and  the  charcoal  was  2  lb.  less  than  \  of 
the  whole :  what  was  the  amount  of  powder  ? 

12.  A  lady  spent  at  one  store  $1  more  than  ^  of  her  money; 
at  another  $^  more  than  ^  of  what  was  left;  at  a  third,  $1 
more  than  -J  of  what  was  left,  and  then  had  $1 :  what  did  she 
have  at  first  ? 

13.  Several  detachments  of  artillery  divided  a  certain  num- 
ber of  cannon-balls :  the  first  took  72  balls  and  -J-  of  the  re- 
mainder; the  second  took  144  balls  and  J  of  the  remainder; 
the  third  took  216  balls  and  ^  of  the  remainder,  and  so  on. 
The  balls  were  thus  divided  equally  among  the  detachments : 
how  many  balls  and  how  many  detachments  ? 

IJ^.  Find  a  number  which,  when  multiplied  by  5,  24  taken 
from  the  product,  the  remainder  divided  by  3,  and  28  taken 
from  the  quotient,  will  give  the  same  number. 

15.  A  and  B  have  the  same  annual  income ;  A  saves  J  of  his, 
but  B  spends  annually  $400  more  than  A ;  at  the  end  of  4  years 
B  finds  himself  $1100  in  debt:    what  is  the  income  of  each  ? 

16.  What  number  is  as  much  more  than  f  as  f  is  less  than  |  ? 

17.  Five  times  |  of  what  number  is  4  less  than  8  times  \ 
of  it? 


ADVANCED    ARITHMETIC.  365 

18.  Seven  times  the  ratio  of  what  number  to  If  is  equal  to 
that  number  added  to  ^  of  the  ratio  of  9  to  4  ? 

19.  Divide  the  number  54  into  two  parts  such  that  f  of  the 
smaller  increased  by  12  will  equal  f  of  the  larger  diminished 
by  3. 

20.  The  sum  of  two  numbers  is  160;  |  of  their  difference  is 
24 :  what  are  the  numbers  ? 

21.  The  greater  of  two  numbers  divided  by  their  difference 
equals  21 ;  their  sum  is  164:  what  are  the  numbers  ? 

22.  One-half  of  the  ratio  of  4  to  6  is  equal  to  ^  of  the  ratio 
of  5  to  what  number  ? 

23.  Twelve  times  the  ratio  of  ^  to  f  is  equal  to  i  of  the  ratio 
of  what  number  to  16  ? 

2Jf..  How  many  times  the  ratio  of  ^  to  |  is  equal  to  3  times 
the  ratio  of  1  to  3  ? 

25.  A  can  do  a  piece  of  work  in  10  days ;  A  and  B  can  do  it 
in  6  days.     In  what  time  can  B  do  it  alone  ? 

26.  A  and  B  can  do  a  piece  of  work  in  16  days ;  after  work- 
ing 4  days,  A  leaves  B  to  finish  the  work,  which  he  does  in  36 
days.     In  what  time  can  each  do  it  alone  ? 

B.    PROBLEMS  OF  TWO  BASES. 

175.  Solving  Equations  of  Two  Unknown 
Numliers, — When  there  are  two  unknown  numbers  to  be 
found,  there  must  be  two  independent  equations  to  be  used  in 
finding  the  unknown  numbers. 

Note.— Two  equations  are  independent  of  each  other,  when  one  can- 
not be  obtained  from  the  other  alone.    Thus, 
.^+2/  =  10  and 

are  independent ;  but 

oc+y  =  10  BLud 
Sx+dy  =  SO, 
are  dependent. 

Two  independent  equations,  relating  the  same  two  unknown 


ADVANCED    ARITHMETIC. 

numbers,  may  be  so  combined  as  to  form  one  new  equation, 
containing  but  one  unknown  number.  This  process  is  called 
Elimination.  There  are  three  plans  of  elimination  :  ( 1 )  by 
substitution,  (2)  by  comparison,  and  (3)  by  addition  and  sub- 
traction. 

I.  By  Substitution. 

EXAMPIiES. 

1.  If  x-\-y  =  10  and  y=4:,  find  the  value  of  x. 

Process:  (1)  x+y  =  10. 
(2)  y  =  4. 
Putting  4  for  the  y  in{l),we  have — 
-      (3).'r+4  =  10. 
Trans.,  (4)  a:  =10-4  =  6,  result. 

2.  2x—y  =  l%  and  5=i.     Find  the  values  of  x  and  y. 

Process:  (1)  2x-y  =  13. 

(2)|  =  i 

9  X  (2)  =  (3)  2/  =  I  =  3,  one  result. 

(1)  =  (4)  2:t'-3=13. 
Trans.,  (5)  2a:  =  13+3  =  16. 
^  of  (5)  =  (6)  x  =  8,  the  other  result. 

S.     x—7y=  —38  and  ^=B.     Find  the  values  of  x  and  y. 
Process  :  (1)  x-7y=  -SS. 

(2)  ^  =  3. 

3x4 

(3)  2/  =  — ^  =  6,  one  result, 

(1)  =  (4)  a:-42=-38. 
Trans.,  (5)  a:  =  42 -38  =  4,  the  other  result. 

4.  x-\-y=lb,  and  x—y—b.     Find  the  value  of  x  and  y. 

Process:  (1)  x+y  =  lb. 
(2)  x-y  =  5. 
Trans.,  (S)  x  =  5  +  y. 

(1)  =  (4)  (5+2/)  +  2/  =  15;  or, 


ADVANCED    ARITHMETIC.  867 

(5)  b  +  y+y=i5. 
Trans,  and  col. ,  (6)  2  ?/  =  15  -  5  =  10. 
(7)  y  =  5, 
(3)  =  (8)  a;  =  5+5  =  10. 

5.  3a:— 42/=10  and  5a;+62/=42.     Find  the  values  of  a;  and  2/. 
Process:  (1)  Sx-^y  =  10. 
(2)  bx+Qy  =  42. 
Trans,  in  (1),  (3)  Sx  =  10+iy. 


iof(3)  = 

=  (4).:= -3    ^ 

(2)  = 

=  (5)  5(1^;^^) +6: 

i/  =  42. 

3X(5): 

=  (6)  5(10+4  ^)  +  18?/ 

=  126. 

(6)^ 

=  (7)  50+202/+18?/  = 

126. 

Trans. 

,  (8)  20i/  +  182/  =  126- 

-50  =  76. 

Ool., 

,  (9)  38^  =  76. 
(10)  2/ =  2. 

(4)  = 

(11)  .  =  ^'.^'  =  6. 

3^_i_7  7x— 6 

6.  -^ — —^  =5  and  ^ — r-^=2.     Find  the  values  of  x  and  y. 
Sy-4:  52/+3 

Process:  (1)  ^^ 1=5. 

32/-4 

7:r-6_ 

^^^5i/+3-2- 

(3  2/-4)x(l)  =  (3)  3.r  +  7  =  152/-20. 

(5 2/ +3) X (2)  =  (4)  7:r-6  =  10*/+6. 

Trans,  in  (3),  (5)  dx-15y=  -27. 

Trans,  in  (4),  (6)  7 a;  =  10 2/ +  12. 


(5)  =  (8)3p±i?]-15.=  -27. 


7x(8)  =  (9)  3(10?/  +  12)-105?/=-189. 
(9)  =  (10)  S0y+SQ-10by=  -189. 
Trans,  and  col.,  (11)  -75 y=  -225. 
(12)  2/ =  8. 

(7)  =  (13)  x  =  ^^  =  6. 


ADVANCED   ARITHMETIC. 


EXERCISE  CLIII. 

Find  the  values  of  the  unknown  numbers :, 


+2/= 22. 


8. 


3. 


3a:-22/  =  10. 


25  a: 


y—x 
8a:=9. 


=  15. 


'\x=y. 


6. 


8 
9a: 


=  -7. 
=  5. 


I  5 


y- 


'  4:X—y 


7.< 


Sx-\-iy     ^' 
^x-4y^^ 
x-y 


ix-S-\-7y=71. 
i     x  —  1 


II.  By  Addition  and  Subtraction. 

EXAMPIiES. 

1.  x+y=25  and  x—y=7.     Find  the  values  of  x  and  y. 

Process:  (1)  x-\-y  =  25. 

(2)  x-y  =  7. 
(l)4-(2)  =  (3)  2 a:  =  32. 

(4)  a:  =  16. 

(1)  =  (5)  16 +  2/ =  25. 

(6)  2/  =  25-16  =  9. 

2.  5a:— 72/=18  and  4a:+32/=23.     Find  the  values  of  a:  and  3/. 

Process:  (1)  6x-7y  =  18. 
(2)  4:X+Sy  =  2S. 
3x(l)  =  (3)  15x-21y  =  b4. 
7x(2)  =  (4)  28.r+21?/  =  161. 
(3)+(4)  =  (5)  43.T  =  215. 
(6)  x  =  5. 

(2)  =  (7)  20+32/  =  23. 
Trans.,  (8)  3?/  =  23-20  =  3. 

(9)2/  =  l. 

Plan  :  (1)  Multiply  the  equations  by  such  numbers  as  will  make  the 


ADVANCED   ARITHMETIC.  869 

coefficient  of  the  x's  or  the  y's  equal.    (2)  If  the  signs  be  unlike,  add;  if 
alike,  subtract. 

Questions:  Why  did  I  nj^ltiply  by  3  and  7  in  No.  (2)?  Could  I  have 
made  the  coefficients  of  the  x's  equal  by  multiplying  by  4  and  5  ?  Can  the 
equations  be  solved  in  that  way  ?  Try  it. 

3.  5  a;— 3  2/ =37  and  6  a;= 11 2/.     Find  the  values  of  a;  and  y. 

Process:  (1)  bx-3y  =  37. 

{2)Qx  =  ny. 
Trans.,  (3)  Qx-lly  =  0. 
5x(3)  =  (4)  30x-5by  =  0. 
6x(l)  =  (5)  30x-lSy  =  222. 
(4)-(5)  =  (6)  -37^= -222. 
(7)  y  =  Q. 
(2)  =  (8)  6:r  =  66. 
(9)a;  =  ll. 

>  4'  T=  5  and  ^ — ,-^=2.     Find  the  values  of  x  and  y. 

Process  :  (1)  ^r -.  =5. 

3y-4i 

■  (2)  ^^^^  =  2. 

(3y-4)x(l)  =  (3)  dx+7  =  15y-20, 

(5i/+3)x(2)=(4)  7x-6  =  10y-\-Q. 
\  Trans,  in  (3),  (5)  3x-15y=  - 27. 

['  Trans,  in  (4).  (6)  7 a; -10 2/ =  12. 

f  2x(5)  =  (7)  6a:-302/=-54. 

I  3x(6)  =  (8)  21x-S0y  =  SQ. 

J  (7)-(8)  =  (9)  -15x=-90. 

[  (10)  x  =  Q. 

(5).=  (11)  18-15?/= -27. 
i  (12)  -15?/= -18-27= -45. 

[  (13)  2/ =  3. 

\ 

t  EXERCISE  CLIV. 

I  Find  the  values  of  x  and  y : 

!  .    {x-\-y  =  29.  '  \Sx-\-y=22. 


370  ADVANCED    ARITHMETIC 

1. 


60. 


5. 


. .  .  8£+5_  . 


4      8       ■  {x+hy      2y+x 

^8+8-'-  |bx-2/=2. 


III.  By  Comparison. 


EXAMPLES. 


1.  x-\-y=12  and  x—y=Q.     Find  the  values  of  x  and  2/. 

Process:   (1)  x+y  =  12. 
(2)  a;-t/  =  6. 
Transpose  in  (1),  (3)  x  =  12-y. 
Transpose  in  (2),  (4)  x  =  Q-\-y. 

(5)  6+2/  =  12-?/.     (Why?) 
Transpose,  (6)  y-\-y  =  12-Q  =  6. 

(7)  22/  =  6. 

(8)  y  =  S. 

(4)  =  (9)  a;  =  6+3  =  9. 

2.  ^x-\-2y  =  17  and  4:X-{-y=lQ.     Find  the  values  of  a:  and  ?/. 

Process;  (1)  3:r+2^  =  17. 
{2)4x+y  =  16. 
Trans,  in  (1),  (3)  2y  =  17-Sx. 

i  of  (3)  =  (4)  2/  =  ^-^. 

Trans,  in  (2),  (b)  y  =  lQ-4x. 

(6)  16-4:r=^^^^'.    (Why?) 

2x(6)  =  (7)  32-8:r  =  17-3a:. 
Trans. ,  (8)  3  a;  - 8  a;  =  17  - 32. 
Col.,  (9)  -5a;=-15. 

(10)  x  =  3. 
(5)  =  (11)  ^  =  16-12  =  4. 

Plan:   Solve  each  equation   for  y  in   terms  of  a*;   then,  make   those 


ADVANCED   ARITHMETIC.  871 

values  of  y  equal.    This  gives  equation  (6),  containing  one  unknown 
number,  x. 

Questions  :  Could  I  have  solved  for  values  of  x  in  terms  of  ^  ?    Would 
this  be  correct  ?    Try  it. 

3.  -^-{—T-=Ssindx—z=—S.     Find  the  values  of  a;  and  z. 

Process:   (1)  4+4  =  8- 
(2)  x-z=-S. 
Trans,  in  (1),  (3)-|=8-^. 

3x(3)  =  (4)  ^  =  24 -?i?. 
4 

Trans,  in  (2),  (5)  x  =  z-B. 

(6)  2-3  =  24-^.    (Why?) 

4x(6)  =  (7)  4^-12  =  96-3^. 

Trans.,  (8)  4 2; +3 2:  =  96 +  12. 
Col.,  (9)  7z  =  108. 

(10)  z  =  15}. 
(5)  =  (11)  a:  =  15f-3  =  12|. 

4.  5x—Sy=S4siud5y—Sx=2.     Find  the  values  of  a;  and  ?/. 

Process  :  (1)  5x-3y  =  M. 
(2)  5i/-3.c  =  2. 
Trans,  in  (1),  (3)  bx  =  S4+Sy. 


Trans,  in  (2),  (5)  -Sx  =  2-by. 
Changing  signs,  (6)  3x  =  by-2. 


(8)^ 

2_S4  +  Sy 

5      •     ^ 

15  X  (8) 

=  (9)  252/- 

10  = 

102+9  y. 

Trans., 

(10)  25  2/- 

9y  = 

=  102  +  10. 

Col., 

(11)  lQy  = 

(12)  y  =  7. 

112. 

(7)  = 

=  (13)  x  =  ^ 

-2 

=  11. 

(Why?) 


872 


ADVANCED    ARITHMETIC. 


5.^  +  ^=4.      '- 

X      y              X  ■ 

r-^ 

Find  the  \ 

Process  . 

-^H 

=  4. 

<- 

1^ 

y 

=  8|. 

Trans,  in  (1) 

<- 

=  4- 

3 

'y' 

^of  (3): 

=<^)^ 

4 
'3 

1 

y' 

Trans,  in  (2) 

iof  (5): 

27yx{7) 
Trans,  and  col. 

<- 

=  (6)  \- 

X 

=  (8)36 
,  (9)  10 

26 

= 

3 

26 
''21 
_l^ 

y 

y- 
y= 

y 

27  =  26y+3. 
30.  1 

(4)  = 

(10)  y-- 
(11)^ 

=  3. 
4 
3 

-h^- 

(12)  X- 

=  1. 

Note. — Instead  of  expressing  the  values  of  x  in  terms  of  y,  I  expressed 
»;^alues  of  -  in  terms  of  y.    This  is  the  more  convenient  plan. 

EXERCISE  CLV. 

Find  the  values  of  x  and  y. 
x-\-Sy=b9. 


'■U: 


=  1 

18. 

„  j4x+2  7/=15. 
'^'\Sx-4y=-S. 


^'  \2x-{-y: 


4- 


x^y      12 

X       y 


6. 


7. 


X     y 

1+1-1 
X     y~lb' 

x—y_Sx—Q 

x—2y=Q. 
nx_6y-22 
5  ""      4      • 
Bx-4:y  =29. 


ADVANCED   ARITHMETIC.  878 

176.   Problems. 

EXAXPIiES. 

1.  The  sum  of  two  numbers  is  50,  their  difference  is  30:  find 

the  numbers. 

Solution  :  Let  a:  =  one  number  and 
2/  =  the  other  number. 
Then,  (1)  x-\-y  =  m 
and  (2)  x-y  =  30. 
(l)+(2)  =  (3)  2:r  =  80. 
(4)  a:  =  40. 
(1)  =  (5)  i0+y  =  50. 
(6)  2/  =  50^40  =  10. 
Answer,  40  and  10. 

2.  A  and  B  together  have  $1500;  A  has  twice  as  much  as  B. 
How  much  has  each  ? 

Solution  :  Let  x  =  A's  money  and 
2/  =  B's  money. 
Then,  (1)  x+y  =  1500 
and  (2)  x  =  2y. 

(1)  =  (3)  2y+y  =  l5O0. 

(4)  3?/ =  1500. 

(5)  2/ =  500. 

(2)  =  (6)  a;  =  1000. 

Answer:  A,  $1000;  and  B,  $500. 

3.  If  I  of  the  time  past  noon  is  equal  to  f  of  the  time  to 
midnight,  what  is  the  hour  ? 

Note. —  The  time  past  noon  plus  the  time  to  midnight  is  12  hours. 

n.  hr.  m. 

past  noon.  to  midnight. 


Solution  :  Let  a:  =  time  past  noon  and 
y  =  time  to  midnight. 
Then,  (1)  x+y  =  12 
and  (2)  |a:  =  f  y. 

|x(2)  =  (3)a:=ii/. 

(i)  =  (4)  hy+y  =  i2. 

(5)  I?/ =  12. 

(6)  y  =  9. 
v3)  =  (7)  .r  =  3. 

As  the  time  past  noon  is  the  hour,  the  answer  is  3  p.  m. 


374  ADVANCED    ARITHMETIC. 

If.  Divide  12  into  two  such  parts  that  3  times  the  one  and  5 
times  the  other  shall  be  46. 

Solution  :  Let  x  =  the  larger  part  and 
2/  =  the  smaller  part. 
Then,  (1)  x+y  =  12 

and  (2)  3x-\-5y  =  4Q. 
3x(l)  =  (3)  3x+Sy  =  3Q. 
(2) -(3)  =  (4)  2  2/ =  10. 
(5)  y  =  5. 

(1)  =  (6)  a:+5  =  12. 

(7)  a:  =  12-5  =  7. 
Answer,  7  and  5. 

5.  A  and  B  had  $30  each ;  after  B  paid  A  a  certain  debt,  A 

then  had  twice  as  much  as  B.     How  much  has  each  ? 
Solution :  Let  x  =  A's  money  at  close  and 
2/  =  B's  money  at  close. 
Then,  (1)  a; +2/ =  60 
and  (2)  x  =  2y. 
(1)  =  (3)  2 2/4-2/ =  60. 

(4)  3  2/ =  60. 

(5)  2/ =  20. 

(1)  =  (6)  a:  +  20  =  60. 
(7)  a:  =  40. 

Answer,  $40  and  $20. 

6.  A  certain  number  is  composed  of  two  digits ;  the  sum  of 
the  units  and  tens  is  6;  and  if  3  times  the  digit  in  tens'  place 
be  subtracted  from  4  times  that  in  units'  place,  the  remainder 
will  he  minus  4.     Find  the  number. 

Solution  :  Let  :r  =  the  units'  figure  and 
2/  =  the  t^n=!'  figure. 
Then  (1)  x-hy=^Q 
and  (2)  4a'-3^v=  -4. 
3x(l)  =  (3)  3.r+3?/  =  18. 

(2)  +  {3)  =  (4)  7.r  =  14. 

(5)  .r  =  2. 

(6)  //  =  6-2  =  4. 
Number,  42. 


ADVANCED    ARITHMETIC.  875 

7.  Thomas  Reed  bought  6%  mining  stock  at  114^%,  and  4% 
furnace  stock  at  112%,  brokerage  i%  in  each  case;  the  latter 
cost  him  $480  more  than  the  former,  but  yields  the  same  annual 
income.     What  did  each  cost  ? 

Note. — The  cost  is  the  market  value  plus  the  brokerage.  Then,  the 
cost  of  the  furnace  stock  is  112 J^,  or  f  of  furnace  stock ;  and  the  cost  of 
the  mining  stock  is  115^,  or  |f  of  mining  stock. 

Solution  :  Let  a:  =  par  value  of  furnace  stock 

and  y  =  par  value  of  the  mining  stock. 
Then  (1)  ^x-Uy  =  ^SO  (why?) 

100x(2)  =  (3)  Qy  =  ix. 
40x(l)  =  (4)  45 .r- 46 2/ =  17200. 
2x(4)  =  (5)  90  rr- 92  2/ =  34400. 
Trans,  in  (3),  (6)   -4x+Qy  =  0. 
22Jx(6)  =  (7)  -Q0x  +  lS5y  =  0. 
(5)+(7)  =  (8)  43  2/ =  34400. 
(9)  y  =  SOO. 
(10)  M2/  =  920. 
(3)  =  (11)  4  a:  =  4800. 

(12)  a;  =1200. 

(13)  ^=1350. 
Answer,  $1350  and  $920. 

8.  5  apples  and  4  oranges  cost  22/,  8  apples  and  7  oranges 
cost  27/.     Find  the  cost  of  1  apj)le  and  1  orange. 

Solution  :  Let  x  =  cost  of  1  apple  and 
y  =  cost  of  1  orange. 
Then,  (1)  5x+iy  =  22  and 

(2)  3  07+ 7  2/ =  27. 
3x(l)  =  (3)  15x  +  12y  =  m. 
5x(2)  =  (4)  15  r +351/ =  135. 
(4)-(3)  =  (5)  23y  =  Q9. 
(6)  y  =  d. 
(2)  =  (7)  3a:+21  =  27. 

(8)  3x  =  Q. 

(9)  a:  =  2. 

Answer,  2f  and  Sf. 


876  ADVANCED    ARITHMETIC. 

9.  A  and  B  can  do  a  piece  of  work  in  16  days ;  after  working 
4  days,  A  leaves  B  to  finish  the  work,  which  he  does  in  86 
days.     In  what  time  can  each  do  it  alone  ? 

Solution  :  Let  a:  =  time  reqd,  by  A  to  do  the  work 
and  y  =  time  reqd.  by  B  to  do  the  work. 

Then,  —  =  part  A  can  do  in  1  da., 
—  =  part  B  can  do  in  1  da., 

and  (2)  1+^+^  =  1.    (Why?) 

4x(2)  =  (3)i5+— =  4. 
X       y 

144 
(3)-(l)  =  (4)  i=  =  3. 

(5)  3j/  =  144. 

(6)  J/  =  48. 

<2)  =  (7)  1+1=1. 

(9)  44. 

(10)  ^  =  24. 

Answer,  24  da.  and  48  da. 

10.  Divide  $600  into  two  parts,  having  the  ratio  of  ^  to  \. 

Solution  :  Let  x  =  larger  part  and 
2/  =  smaller  part. 
Then,  (1)  x+y-=600 
and  {2)  x:y::l:l. 

(2)  =  (3)^=|. 

2/X(3)  =  (4)^=?|. 
Trans,  in  (1),  (5)  x  =  Q0O-y. 

{Q)^  =  QOO-y.    (Why?) 


ADVANCED   ARITHMETIC.  377 

(7)  Qy  =  2400-4y. 

(8)  61/ +4  2/ =  2400. 

(9)  10i/  =  2400. 
(10)  y  =  240. 

(4)  =  (ll)a:=^^  =  360. 

Answer,  $360  and  $240. 

11.  At  what  time  between  4  and  5  o'clock  are  the  hour-hand 
and  minute-hand  of  a  clock  together  ? 

Note. — At  4  o'clock  the  minute-hand  is  at  XII  and  the  hour-hand  is  at 
nil.  Then,  when  the  minute-hand  gains  20  minute  spaces  on  the  hour- 
hand  they  will  be  together. 

Solution:  Let  .r  =  the  minute-hand's  travel 

and  y  =  the  hour-hand's  travel. 
Then,  (1)  sc-y  =  20 
and  (2)  x  =  12y.    (Why?) 
Trans,  in  (1),  (3)  x  =  20+y. 

(4)  12y  =  20  +  y.    (Why?) 

(5)  12y-y  =  20. 

(6)  111/ =  20. 

(7)  2/  =  !?. 

(8)  .  =  ^  =  21^, 
Answer,  21  r\  min.  after  4  o'clock. 

12.  At  what  time  between  4  and  5  o'clock  are  the  hour-hand 
and  the  minute-hand  equally  distant  from  V  ? 

Note.— There  are  two  answers  to  this  problem:  (1)  When  the  two 
hands  are  together,  the  conditions  are  the  same  as  in  Example  4;  (2) 
when  the  hands  are  on  opposite  sides  of  V.  We  will  solve  for  the  second 
answer. 

Draw  a  clock-face  and  verify  this  statement :  When  the  hands  reach 
the  required  places,  the  minute-hand  will  lack  as  far  of  being  to  VI  as 
the  hour-hand  is  past  IIII. 

Solution  :  Let  x  =  min. -hand's  travel  (4)  13  y  =  30. 

and  y  =  hour-hand's  travel.  ,_>      _  30 

Then,  (1)  a:  =  30 -2/     (statement  ^^^  ^  ~  l3' 

above)  _ /^n  ^ _  12x30 _  _  , 

And  (2)  x  =  12y.    (Why?)  {^)-{G)  x-     ^^     -zt^^. 

(3)  12^  =  30-?/.    (Why  ?)       Answ^er,  27tV  min.  after  4  o'clock. 


878  ADVANCED   ARITHMETIC. 

13.  The  head  of  a  fish  weighs  9  pounds,  the  tail  weighs  as 
much  as  the  head  and  half  the  body,  and  the  body  weighs  as 
much  as  the  head  and  tail  together.  Find  the  weight  of  the 
fish. 

Solution :  Let  x  =  weight  of  body 
and  2/  =  weight  of  tail. 
Then,  (1)  2/  =  9  +  ia: 
and  (2)  x  =  y-]-9. 
Trans,  in  (2)  (3)  y  =  x-9. 

{4)x-d  =  9  +  lx.    (Why?) 
Trans.,  (5)  a:-ia:  =  9+9  =  18. 

(6)  :r  =  36.  ' 

(7)  2/  =  36-9  =  27. 

Answer  =  36  lb. +27  lb. +9  lb.  =  72  lb. 

14'  A  and  B  rent  a  field  for  $27 ;  A  puts  in  4  horses  for  5 
months,  and  B  puts  in  10  cows  for  6  months :  what  ought  each 
to  pay  if  2  horses  eat  as  much  as  8  cows  ? 

Solution  :  (1)  Pasture  for  2  h.  =  pasture  for  3  c. 
2  X  (1)  =  (2)  Pasture  for  4  h.  =  pasture  for  6  c. 

.'.  A's  4  horses  are  equivalent  to  6  cows. 

(3)  Pasture  of  6  c.  for  5  mo,  =  pasture  of  30  c.  for  1  mo. 

(4)  Pasture  of  10  c.  for  6  mo.  =  pasture  of  60  c.  for  1  mo. 
.'.  they  should  pay  in  the  ratio  of  30  to  60. 

Let  x  =  A's  part  and 
2/  =  B's  part. 
Then,  (5)  x+y  =  27 
and  (6)  x:y::30:Q0. 

m  =  i7)j  =  U  =  h 

(8)  2x  =  y. 
(5)  =  (9)  a:+2rr  =  27. 

(10)  x  =  9. 

(11)  2/ =  18. 

Answer,  $9  and  $18. 

15.  A  fox  is  70  leaps  in  advance  of  a  hound ;  the  fox  takes  8 
leaps  to  the  hound's  6,  but  2  of  the  hound's  leaps  equal  5  of 


ADVANCED    ARITHMETIC.  379 

the  fox's:  how  many  leaps  must  the  hound  take  to  catch  the 

fox? 

Solution:  (1)  5f.L  =  2h.l. 
iof(l)  =  (2)  lf.l.  =  |h.l. 
8x(2)  =  (3)  8f.l.  =  3ih.l. 
70x(2)  =  (4)  70f.l.=28h.l. 

Note.— Re-writing  the  problem,  putting  for  fox-leaps  their  equiva- 
lents found  in  (3)  and  (4),  the  leaps  will  all  be  of  the  same  length.  We 
have,  **A  fox  is  28  leaps  ahead  of  a  hound,  and  takes  3^  leaps  while  the 
hound  takes  6.  How  many  leaps  will  the  hound  make  in  overtaking  the 
fox,  the  leaps  being  all  of  the  same  length  ?  " 

8v 
Let  a:  =  No.  of  leaps  for  the  (8)  ^=  j^g- 

^         ^^''^^  f      .y.         Trans.in(5),(9):r  =  i/-28. 

and  2/  =  No.  of  leaps  for  the  o 

hound.  (10)  y-28  =  ^. 

Then,  (5)  y-x  =  2S  15x(10)  =  (ll)  lby-420  =  Sy. 

and  {Q)x:y::Sl:6.  Trans.,  (12)  Iby-Sy  =  i20. 

(6)  =  (7)  ^  =  U  =  j\-  (13)  ?/  =  60. 

^  Answer,  60  leaps. 

Note. — If  fox-leaps  had  been  required  in  the  answer,  the  hound-leaps 
should  have  been  reduced  to  fox-leaps. 

16.  A  banker  owns  2|%  stocks  bought  at  10%  below  par, 
and  8%  stocks  bought  at  15%  below  par.  The  income  from  the 
former  is  66f  %  more  than  from  the  latter,  and  the  investment 
in  the  latter  is  $11400  less  than  in  the  former.  Find  the  invest- 
ment in  each. 

Solution  :  Let  x  =  par  value  of  1st  (3)  -5-  2|  =  (5)  x  =  2y. 

stocks.  (6)  180  2/ -85?/ =  1140000. 

and  y  =  par  value  of  2d  (Why  ?) 

stocks.  (7)  951/ =  1140000. 

(Why?)  ^^^100"^^^- 

100      100         ^^^j^y^^  (11)  ?^  =  21600. 


100x(l)  =  (3)  2lx  =  6y. 

100  X  (2)  =  (4)  90  X  -  85  //  =  1140000. 


100 
Answer,  $10200  and  $21600. 


380  ADVANCED   ARITHMETIC. 

17.  My  agent  sold  my  flour  at  4%  commission;  increasing 
the  proceeds  by  $4.20,  I  ordered  the  purchase  of  wheat  at  2% 
commission ;  after  which,  wheat  declining  8^%,  my  whole  loss 
was  $5.     What  was  the  selling  price  of  the  flour  ? 

Solution  :  Let  a:  =  selling  price  of  flour 

and  y  =  cost  price  of  wheat. 

96  X 
Then,  —^= proceeds  from  flour, 

QQX 

r—  +4.2  =  investment  in  wheat, 

and(2)  ??||  =  :r  +  4.2-5.    (Why?) 

100x(l)  =  (3)  102?/  =  96.r  +  420. 
3(X)x(2)  =  (4)  290?/ =  300 a- +1260 -1500. 
(5)  29i/  =  30.r-24. 
Jof  (3)  =  (6)  342/  =  32a:+140. 
16x(5)  =  (7)  464 2/ =  480^-384. 
15x(6)  =  (8)  510i/  =  480.r+2100. 
(8) -(7)  =  (9)  46  2/ =  2484. 
(10)  2/ =  54. 

Trans,  in  (2),  (11)  x  =  ^-^+5-4.2. 

(12)ar  =  ^5|^+5-4.2  =  $53. 

18.  Suppose  10%  state  stock  20%  better  in  market  than  4% 
railroad  stock.  If  A's  income  be  $500  from  each,  how  much 
has  he  paid  for  each,  the  whole  investment  bringing  6^f3%  ? 

Note. — The  solution  to  this  problem  is  given  in  four  steps :  (1)  We  ob- 
tain the  par  value  of  the  state  stock ;  (2)  the  par  value  of  the  railroad 
stock ;  (3)  the  amount  of  the  investment ;  and  (4)  with  the  results  ob- 
tained from  the  first  three  steps  we  are  enabled  to  obtain  the  bases  for 
the  last  step,  which,  when  solved,  gives  required  results. 

Step  1 :  (1)  W  s.  s.  =  $500,  income  on  s.  s.    (Basis.) 
10x(l)  =  (2)  100^  s.  s.  =  $5000,  par  value  of  s.  s. 
Step  2  :  (1)  4%  r.  r.  s.  =$500,  income  on  r.  r.  s.     (Basis.) 
25x(l)  =  (2)  100^  r.r.  s.  =  $12500,  par  value  of  r.  r.  s. 


ADVANCED    ARITHMETIC.  881 

Step  3  :  (1)  .Qsh[%  inv.  =  $1000,  whole  income.     (Basis.) 
i^\%  of  (1)  =  (2)  1%  inv.  =  $166.50. 
100  X  (2)  =  (3)  100^  inv.  =$16650. 

Note.— There  are  50  shares  state  stock  and  125  shares  r.  r.  stock. 

Step  4  :  Let  a:  =  market  value  of  1  share  s.  s. 

and  ^  =  market  value  of  1  share  r.  r.  s. 
Then,  (1)  50^  =  ^^x50?/ (why  ?) 
and  (2)  50  a: +  1251/ =  16650. 
Trans,  in  (1),  (3)  5Ox-Q0y  =  0. 
(2) -(3)  =  (4)  185^  =  16650. 

(5)  y  =  90. 

(6)  5  a:  =  540. 

(7)  a;  =  108. 

State  stock:  50  sh.  at  108^  =  $5400,  answer. 
R.  R.  stock:  125  sh.  at  90?^  =  $11250,  answer. 

EXERCISE  GLVI. 

1.  Divide  $500  between  A  and  B,  so  that  A  may  have  $40 
more  than  B. 

2.  Divide  $420  between  Charles  and  Henry,  so  that  Henry 
will  have  $25  less  than  Charles. 

3.  Two  men  owe  a  debt  of  $54  in  the  ratio  of  4  to  5 :  what 
does  each  owe  ? 

4.  Two  men  owe  a  debt;  for  every  dollar  that  A  owes,  B 
owes  $1.50.     If  the  debt  is  $250,  find  each  man's  part. 

5.  Divide  20  into  two  such  parts  that  the  smaller  plus  5  shall 
equal  the  larger  minus  5. 

6.  A  and  B  have  $500;  A  says  to  B,  "  Give  me  $50  and  then 
I  will  have  as  much  as  you  will  have."     How  much  has  each  ? 

7.  James  and  John  bought  a  v/atermelon  for  $.85 ;  they  di- 
vided it  so  that  John  got  2|  times  as  much  as  James :  what 
should  each  pay  ? 

8.  i  of  the  time  past  noon  is  equal  to  ^  of  the  time  to  mid- 
night :    what  is  the  time  ? 


OH^'^^' 


382  ADVANCED    ARITHMETIC. 

9.  i  of  the  time  past  noon  plus  8^  hours  is  equal  to  |  of  the 
time  to  midnight.     Find  the  hour. 

10.  -J  of  the  time  past  midnight  plus  2  hours  is  equal  to  ^  of 
the  time  to  noon  minus  1^  hours :  find  the  time  past  midnight. 

11.  All  the  time  past  midnight  minus  2  hours  is  equal  to  \ 
of  the  time  to  noon.     What  is  the  time  ? 

12.  A  number  is  composed  of  2  digits ;  the  sum  of  the  digits 
is  5,  and  the  number  of  units  is  f  of  the  number  of  tens.  Find 
the  number. 

13.  A  number  is  composed  of  2  digits ;  the  sum  of  the  digits  is 
12 ;  5  times  the  number  of  units  is  equal  to  2^  times  the  num- 
ber of  tens.     Find  the  number. 

IJf..  A  number  has  two  places ;  the  sum  of  the  units  and  tens 
is  4;  if  86  be  subtracted,  the  order  will  be  reversed.  Find  the 
number. 

Note. — Any  number  expressed  by  the  French  system  of  notation  is 
equal  to  the  sum  of  its  1st  ( units' )  digit,  10  times  its  2d  ( tens' )  digit,  100 
times  its  3d  (hundreds')  digit,  and  so  on.  Convince  yourself  of  this 
truth  by  applymg  it  to  several  numbers. 

15.  A  and  B  are  partners ;  A  invests  a  capital  of  $500,  and  B 
a  capital  of  $850;  the  gain  is  $540:  divide  it. 

16.  A  and  B  are  partners ;  they  invested  capital  in  the  ratio 
of  5  to  8 ;  they  have  gained  $650 :  divide  it. 

17.  Divide  i  in  the  ratio  of  6  to  7. 

18.  Divide  .06  in  the  ratio  of  ^  to  i. 

19.  Divide  10.6  in  the  ratio  of  .1  and  .05. 

20.  A  and  B  are  partners ;  A  puts  in  $660  for  10  months, 
and  B  has  in  $500  for  the  first  6  months  and  only  $100  for 
the  other  4  months ;  they  gain  $600 :  divide  it. 

21.  A  and  B  have  a  joint  stock  of  $2000  by  which  they  gain 
$640,  of  which  A  receives  $128  more  than  B:  what  is  each 
man's  share  of  the  stock  ? 

22.  How  many  minutes  does  it  lack  of  4  o'clock,  if  |  of  an 
hour  ago  it  was  twice  as  many  minutes  past  2  o'clock  ? 


ADVANCED    ARITHMETIC.  383 

23.  A  and  B  start  at  the  same  time  from  two  places,  M  and 
N,  154  miles  apart,  and  each  travels  toward  the  other  till  they 
meet ;  A  travels  3  miles  in  2  hours,  and  B  travels  5  miles  in  4 
hours  :  where  will  they  meet  ? 

Note. — In  1  hour  A  travels  1|  miles  and  B  travels  IJ  miles ;  therefore 
the  distance  is  to  be  divided  in  the  ratio  of  1|  to  1^. 

24..  A  house  and  garden  cost  $850;  5  times  the  price  of  the 
house  equals  12  times  the  cost  of  the  garden  :  find  the  cost  of 
each. 

25.  The  sum  of  two  numbers  is  5760,  and  their  difference  is 
equal  to  i  of  the  greater:  find  the  numbers. 

26.  A  agrees  to  work  for  $2  a  day  and  forfeit  $1  for  every 
day  he  is  idle ;  at  the  end  of  20  days  he  receives  $25 :  how  many 
days  does  he  idle  ? 

Note. — To  receive  $25  he  must  work  12|  days.  During  the  remaining 
7i  days  he  forfeited  all  he  earned  ;  therefore  he  idled  twice  as  much  as 
he  worked.  The  problem  then  becomes:  Divide  7^  into  two  parts  in  the 
ratio  of  1  to  2. 

27.  A  has  7  loaves  of  bread,  B  5,  and  C  none.  The  three  eat 
all  of  the  bread,  each  the  same  amount;  C  pays  to  A  and  B  12 
cents  :  what  shoilld  each  receive  ? 

Note.— Each  will  eat  one-third  of  the  twelve  loaves,  or  4  loaves.  A 
eats  4,  and  has  3  left  for  C  ;  B  eats  4,  and  has  1  left  for  0.  C  therefore 
eats  3  of  A's  and  1  of  B's,  and  the  12  cents  should  be  divided  in  the  ratio 
of  3  to  1. 

28.  Two  boys  run  a  race :  the  smaller  boy  steps  4  feet  and  the 
larger  steps  6  feet;  and  the  smaller  boy  takes  5  steps  while  the 
larger  boy  takes  4;  the  larger  boy  gives  the  smaller  120  feet  the 
start,  and  they  come  out  even  :  how  many  steps  does  the  larger 
boy  take,  and  how  far  does  he  run  ? 

29.  In  the  last  number,  how  many  steps  does  the  smaller  boy 
take,  and  how  far  does  he  run  ? 


884  ADVANCED   ARITHMETIC. 

30.  Two  cog-wheels  work  together;  one  has  11  cogs  and  the 
other  85  cogs :  in  how  many  rounds  of  the  large  wheel  will  the 
smaller  gain  72  rounds  ? 

81.  A  hare  is  80  leaps  in  advance  of  a  greyhound ;  the  hare 
makes  11  leaps  while  the  hound  makes  9,  and  leajis  four-fifths 
as  far  as  the  hound :  how  many  more  leaps  will  the  hare  make 
before  the  hound  catches  him  ? 

32.  There  are  two  numbers,  the  one  twice  as  large  as  the 
other ;  one-third  of  the  smaller  and  one-half  of  the  larger  equal 
20 :  find  the  numbers. 

33.  Four-fifths  of  A's  money  is  equal  to  one-half  of  B's;  A 
gains  $200  and  B  loses  $100;  they  now  have  the  same:  what 
had  they  at  first  ? 

31/..  A  number  is  composed  of  2  digits;  the  number  is  equal 
to  9  times  the  units  less  18 ;  it  is  also  equal  to  12  times  the 
difference  between  the  units  and  the  tens :  find  the  number. 

35.  A  number  is  composed  of  two  digits :  the  difference  be- 
tween the  units  and  the  tens  is  4 ;  5  times  the  units  is  equal  to 
2^  times  the  tens :  find  the  number. 

36.  A  is  4  times  as  old  as  his  son ;  in  14  years  he  will  be  only 
twice  as  old  as  the  son :  find  their  ages. 

37.  At  what  time  between  8  o'clock  and  9  o'clock  are  the 
hands  of  a  clock  together  ? 

38.  At  what  time  between  8  o'clock  and  9  o'clock  are  the 
hands  of  a  clock  opposite  each  other  ? 

39.  At  what  times  between  5  and  6  is  the  minute-hand  half- 
way between  the  hour-hand  and  XII  ? 

Ji.0.  At  what  times  between  5  and  6  will  the  hands  be  at  right 
angles  to  each  other  ? 

j^l.  If  one-half  of  the  time  past  midnight  is  equal  to  one- 
sixth  of  the  time  past  noon,  what  is  the  hour  ? 

4.2.  If  one-fourth  of  the  time  past  noon  is  equal  to  one-six- 
teenth of  the  time  past  midnight,  what  is  the  hour  ? 


ADVANCED    ARITHMETIC.  385 

4.3.  If  the  time  to  noon  equals  one-seventh  of  the  time  to 
midnight,  what  is  the  hour  ? 

4.4..  If  one-half  of  the  time  to  midnight  is  equal  to  one- 
eighth  of  the  time  to  noon,  what  is  the  hour  ? 

4.5.  A  farmer  sold  to  one  man  20  bushels  of  wheat  and  8 
bushels  of  oats  for  $14 ;  to  another  12  bushels  of  wheat  and  10 
bushels  of  oats  for  $9.70:  find  the  price  of  the  wheat  and  the 
oats. 

46.  A  man  bought  8  cows  and  7  calves  for  $95;  again,  he 
bought  5  cows  and  4  calves  for  $120 :  find  the  price  of  a  cow 
and  a  calf. 

47.  A  is  30  years  old;  A's  age  is  equal  to  B's  plus  one-half 
of  C's ;  and  C  is  as  old  as  A  and  B :  find  the  ages  of  B  and  C. 

4.8.  The  head  of  a  fish  is  6  inches  long ;  its  tail  is  as  long  as 
the  head  and  one-fourth  of  the  body ;  and  the  body  is  as  long 
as  the  head  and  the  tail :  find  the  length  of  the  fish. 

49.  A  bag  contains  three  times  as  many  dollars  as  quarters ; 
if  8  dollars  and  8  quarters  be  taken  away,  there  will  be  5  times 
as  many  dollars  as  quarters :  find  the  number  of  each. 

50.  A  has  two  horses  and  a  saddle ;  the  saddle  is  worth  $20 ; 
if  it  be  placed  on  one  horse  it  will  make  him  worth  as  much  as 
the  other ;  but  if  it  be  placed  on  the  second  horse,  he  is  then 
worth  twice  as  much  as  the  first :  find  the  value  of  each  horse. 

51.  A  flag-pole  consists  of  two  parts ;  the  length  of  the  upper 
is  five-sevenths  of  the  length  of  the  lower;  and  9  times  the 
upper  part  added  to  13  times  the  lower  part  is  longer  than  11 
times  the  whole  pole  by  36  feet ;  find  the  length  of  the  pole.     - 

52.  A  party  was  composed  of  men  and  women;  6  of  the 
women  left ;  there  were  then  twice  as  many  men  as  women ; 
when  the  6  women  returned  with  their  husbands,  the  number 
of  women  was  only  two-thirds  of  the  number  of  men.  What 
was  the  size  of  the  original  party  ? 


386  ADVANCED   ARITHMETIC. 

53.  Twenty-six  tons  can  be  carried  by  15  wagons  and  22 
carts,  or  by  18  wagons  and  16  carts :  what  is  a  load  for  a  cart 
and  for  a  wagon  ? 

54..  A  and  B  started  from  M  and  N  respectively,  and  trav- 
eled till  they  met ;  it  appeared  that  A  had  traveled  five-sev- 
enths as  far  as  B ;  but  if  A  had  traveled  15  miles  farther,  he 
would  have  traveled  twice  as  far  as  B  :  how  far  did  each  travel? 

55.  A  cistern  has  two  pipes ;  one  can  fill  it  in  15  hours  and 
the  other  can  empty  it  in  18  hours :  if  both  are  left  open,  what 
time  will  the  cistern  be  in  filling  ? 

56.  A  cistern  of  860  gallons  has  two  pipes ;  one  can  fill  it  in 
15  hours  and  the  other  can  empty  it  in  20  hours :  if  both  are 
left  open,  what  time  will  the  cistern  be  in  filling  ? 

57.  A,  after  doing  :i  of  a  work  in  5  days,  calls  the  assist- 
ance of  B,  and  they  finish  the  work  in  6  days:  in  what  time 
could  each  have  done  the  whole  work  ? 

58.  A  and  B  could  have  done  a  work  in  15  days,  but  after 
working  together  6  days,  B  was  left  to  finish  it,  which  he  did 
in  80  days :  in  what  time  could  A  have  finished  it  if  B  had  left 
at  the  end  of  the  6  days  ? 

C.     PROBLEMS  OF  THREE  OR  MORE  BASES. 

177.  Solving  Equations  of  Tliree  or  More 
Unknown  Numbers. — When  there  are  three  or  more  un- 
known numbers  to  be  found,  there  must  be  as  many  equations 
(bases)  as  there  are  unknown  numbers. 

The  process  of  elimination  here  is  not  different  from  that 
given  under  Problems  of  Two  Bases. 


ADVANCED    ARITHMETIC.  387 

EXAMFIjES. 

/.  Sx-i-^y-z  =  lS,    6x-Qy-\-5z  =  S,    x-\-y-^z  =  Q.      Find    the 
values  of  x,  y^  and  z. 

Process:  (1)  Sx+by-z  =  lS. 

(2)  bx-6y+5z  =  8. 

(3)  x+y+z  =  Q. 
3x(3)  =  (4)  Sx-{-Sy-^Sz  =  l8. 

(l)-(4)  =  (5)22/-4s  =  0. 

5x(3)  =  (6)  6x+by+5z  =  30. 
(6) -(2)  =  (7)  11 2/ =  22. 
(8)  y  =  2. 
(5)  =  (9)  4-4^  =  0. 

(10)  z  =  l. 
(3)  =  (11)  x-\-2+l  =  Q, 
(12)  a;  =  3. 

^.  x-^y-{-z=22,  y-\-z-{-r=21,  x+z-^r=ld,  x-\-y-^r=m.  Find 
the  values  of  x,  y,  z,  and  r. 

Process:  (!)  x-\-y-\-z  =  22. 

(2)  y  +  2:+r  =  21. 

(3)  x+z+r  =  19. 

(4)  .T+y+r  =  16. 
Adding,  (5)  3x+3?/+32+3r  =  78. 

Jof  (5)  =  (6)  ^+2/+2;+r  =  26. 
(6)-(I)  =  (7)  r  =  4. 
(6) -(2)  =  (8)  :r  =  5. 
(6) -(3)  =  (9)  2/ =  7. 
(6) -(4)  =  (10)  2  =  10. 

3.  Sx+2y=lS,  Sy-2z  =  S,  2^-32=9.     Find  the  values  of 

x^  y,  and  z. 

Process:  (1)  2x+2y  =  ld.  9x(3)  =  (8)  18 a; -27 2  =  81. 

{2)Sy-2z  =  8.  (8)-(7)  =  (9)   -35^  =  35. 

(3)  2.^-3^  =  9.  (10)  z=-l. 

3x(l)  =  (4)  9:c  +  6^  =  39.  (2)  =  (11)  32/+2  =  8. 

2x(2)  =  (5)  62/-40  =  16.  (12)  ?/  =  2. 

(4)-(5)  =  (6)  9.C  +  42+23.  (1)  =  (13)  3^+4  =  13. 

2x(6)  =  (7)  18^+8^  =  46.  (14)  x  =  S. 


888 


ADVANCED    ARITHMETIC. 


EXERCISE  CLVII. 
Find  the  values  of  the  unknown  numbers  : 


5. 


{  2x-y-\-4:Z=2Q. 
i  Qx-z  =  -14:. 

{x-{-Sy-\-2z  =  17. 
]4x-2y-\-z  =  12. 
(Sx-y  =  10. 

{Qx+by-\-4:Z=S7. 
■{  7x-Sy-\-dz  =  2Q. 
(Sx-{-2y-z=S. 

x^y+z  =  7. 
y-\-z-{-w=Q. 
z-\-w-\-x=S. 
iv-\-x-{-y=9. 


6. 


■1 


^x-\-y  =  n. 

72+5r=:ll. 
[3r+2x=4. 
8a;-102/+-ls=60. 
7a;+3  2/-52=23. 
x-2y-\-4:Z=m. 

6  "^  8  "^  3 


9. 


2;+2/+2=43. 


178.   Problems. 

EXAMPLES. 

1.  A,  B  and  C  have  $210:  twice  A's  money  and  3  times  B's 
money  added  to  C's  will  make  $400;  A's  money  and  4  times 
B's  added  to  twice  C's  will  make  $510.     How  much  has  each  ? 
Solution  :  Let  r  =  A's  money, 

//  =  B's  money,  and 
2  =  C's  money. 
Then,  (1)  x+y+z  =  210, 

(2)  2x-hSij+z  =  400, 
and  (3)  x+iy-\-2z  =  bl0. 
(3)-(l)  =  (4)  Sy-^z  =  SOO. 

2x(l)  =  (5)  2x+2y-\-2z  =  420. 
(5)-(2)  =  {6)   -y+z  =  20. 
(4) -(6)  =  (7)  41/ =  280. 
(8)  2/ =  70. 
(6)  =  (9)  -70+2  =  20. 

(10)  2  =  90. 
(1)  =  (11)  .t'+70  +  90  =  210. 
(12)  .1=50. 

Answer,  .$50,  $70,  and  $90. 


ADVANCED   ARITHMETIC.  389 

2.  A  and  B  together  have  $600;  A  and  C,  $700;  B  and  C, 
$500.     How  much  has  each  ? 

Solution:  Let  :r  =  A's  money, 

2/  =  B's  money,  and 
2;  =  G's  money. 
Then,  (1)  x+y  =  Q0O, 
(2)  x+z  =  700, 
and  (3)  y+z  =  bOO. 
Adding  (4)  2.r+2?/+2s  =  1800. 

(5)  x+y-\-z  =  900. 
(5)-(l)  =  (6)  s  =  300. 
(5) -(2)  =  (7)  2/ =  200. 
(5) -(3)  =  (8)  a:  =  400. 

Answer,  $400,  $200,  and  $300. 

3.  A  number  is  expressed  by  3.  digits,  the  sum  of  which  is  9; 
the  number  is  42  times  the  sum  of  the  third  and  the  second 
digits,  and  the  first  (units')  digit  is  twice  the  sum  of  the  other 
two :  find  the  number. 

Note. — Any  number  expressed  by  the  French  system  of  notation  is 
equal  to  the  sum  of  its  1st  (units')  digit,  10  times  its  2d  (tens')  digit,  100 
times  its  3d  (hundreds')  digit,  and   so  on.     Convince  yourself  of  this 
truth  by  applying  it  to  several  numbers. 

Solution:  Let.r  =  units*  figure.       Trans,  in  (3),  (7)  x-2y-2z  =  0. 

2/  =  tens' figure,  (i)-(7)  =  (8)  Sy+Sz  =  9. 

and  2  =  hundreds'  (9)  y+z  =  S. 

figure.  ^  of  (6)  =  (10)  19z-ny=-S. 

Then,(l)  x-{-y+z  =  9,  llx(9)  =  (ll)  llz+ny  =  3S. 

(2)  100z-{-10y-\-x=  (10)  +  (11)  =  (12)  30^  =  30. 

42(2/+2),  (13)  z  =  l. 

and  (3)  x=:2{z+y)  =  2z+  (9)  =  (14)  y+l  =  S. 

^y-  (15)  y  =  2, 

(2)  =  (4)  100z+10y-{-x=  (1)  =  (16)  a:+2+l  =  9. 

42^  +  420.  (17)  a:  =  6. 

Col.,  (5)  r-32y+b8z  =  0.  Number,  126. 
(5)-(l)  =  (6)  57^-33^= -9. 


390  ADVANCED   ARITHMETIC. 

^.  A,  B,  C,  D  and  E  have  money;  B  gives  A  :|  of  his ;  C  gives 
B  ^  of  his ;  D  gives  C  i  of  his ;  and  E  gives  D  ^  of  his :  each 
then  has  $80.     How  much  had  each  at  first  ? 
Solution  :  Let  x  =  A's  money,  (4)  =  (7)  |  w; +6  =  30. 

y  =  B's  money,  (8)  f  ly  =  24. 

3  =  0'smoney,  (9)  w  =  S2. 

w  =  I>'s  money,  and  (3)  =  (10)  f  s  +  8  =  30. 

r  =  E's  money.  (11)3  =  33. 

Then,  (1)  a:+i|/  =  30  (2)  =  (12)  i?/  +  ll  =  30. 

(2)  ^2/+i2  =  30,  (13)  y  =  SS. 

(3)  fs  +  it/;  =  30,  (1)  =  (14)  rr+19  =  30. 

(4)  |^/;  +  ir  =  30,  (15)  x  =  ll. 

and  (5)  f  r  =  30.  Answer,  $11,  $38,  $33,  $32,  $36. 

gof  (5)  =  (6)  r  =  36. 

5.  A  and  B  can  do  a  piece  of  work  in  12  days ;  A  and  C  in 
15  days ;  and  B  and  C  in  20  days.     In  what  time  can  each  do  it 

alone  ? 

Solution :  Let  r  =  time  req'd  by  A  to  do  the 
work, 
y  =  time  req'd  by  B  to  do  the 
work, 
and  2  =  time  req'd  by  C  to  do  the 
work. 

Then,  — =part  done  by  A  in  1  da. 

—  =  part  done  by  B  in  1  da. 
2/ 

—  =  part  done  by  C  in  1  da. 


(1) 

'> 

12 

=1, 

(2) 

X 

15 

z 

=1, 

and  (3) 

20^ 

y 

20 

s 

=  1. 

-iVof(2)  = 

=  (4) 

1  + 

X 

1 

z 

1 

15* 

hofiS)-- 

=  (5) 

y 

1 

z 

1 
20' 

I 


ADVANCED    ARITHMETIC.  391 


Adding,  (7)  A+A+A  =  ^. 

^of7  =  (8)4+^+^  =  A  =  _^ 

(8)-(6)  =  (9)^  =  -i,and2  =  60. 

(8) -(4)  =  (10)  -^=-^,  and  y  =  dO. 

(8)-(5)  =  (ll)^  =  ^,anda:  =  20. 
Answer,  20  da.,  30  da.,  and  60  da. 


EXERCISE  OLVIII. 

1.  Two  calves,  5  sheep  and  8  hogs  cost  $84;  8  calves,  6  sheep 
and  7  hogs  cost  $108 ;  10  calves,  12  sheep  and  8  hogs  cost  $98 : 
find  the  cost  of  a  calf,  a'  sheep,  and  a  hog. 

2.  I  bought  corn,  wheat  and  oats,  as  follows:  4  bushels  of 
corn  and  3  bushels  of  wheat  for  $3.70;  6  bushels  of  wheat  and 
8  bushels  of  oats  for  $6.20;  12  bushels  of  corn  and  9  bushels 
of  oats  for  $7.05:  find  the  price  of  each  per  bushel. 

3.  A,  B  and  C  have  $1000;  A  and  B  together  own  $800;  B 
and  C  together  own  $630 :  what  does  each  own  ? 

4..  A,  B  and  C  have  $3600;  if  B  gives  A  one-half  of  his,  and 
C  gives  B  one-third  of  his,  they  will  all  have  the  same  amount : 
what  had  each  at  first  ? 

5.  After  a  battle  in  which  24000  men  were  engaged,  it  was 
found  that  the  number  slain  was  one-seventh  of  those  who  sur- 
vived, and  that  the  number  wounded  was  equal  to  one-half  of 
the  slain  :  find  the  number  slain,  the  number  wounded,  and  the 
number  not  hurt. 

6.  A  farmer  has  45  head  of  horses,  cows,  and  sheep;  there 
are  six  times  as  many  sheep  as  cows,  and  one-third  as  many 
horses  as  sheep :    find  the  number  of  each. 


892  ADVANCED    ARITHMETIC. 

7.  A,  B  and  C  built  a  wall  200  feet  long ;  B  built  as  many 
feet  as  C  and  one  third  more,  and  A  built  three-fourths  as 
much  as  B  :    how  many  feet  did  each  build  ? 

8.  A  man  divided  his  estate  of  $4000  as  follows :  To  his  wife 
$500  more  than  to  his  son,  and  to  the  son  $1000  more  than  to 
the  daughter:  find  the  share  of  each. 

9.  Four  persons  compare  their  money ;  the  first  man  has  one- 
half  as  much  as  all  the  others ;  the  second,  one  third  as  much 
as  all  the  others ;  the  third,  one  fourth  as  much  as  the  other 
three;  and  the  fourth,  $14  less  than  the  first:  find  the  amount 
of  each  man's  money. 

10.  A  says  to  B,  "  Give  me  $100  and  I  will  have  as  much 
money  as  you  will  have;"  and  he  says  to  C,  "  Give  me  $200 
and  I  will  have  as  much  as  you  will  have."  B  says  to  C,  "Give 
me  $350  and  my  money  will  be  to  yours  as  19:  9."  Find  how 
much  each  has. 

11.  A  and  B  can  do  a  piece  of  work  in  48  days ;  A  and  C  can 
do  it  in  80  days ;  B  and  C  can  do  it  in  26f  days :  in  what  time 
can  each  do  it  alone  ? 

12.  A,  B  and  C  earn  $68  in  14  days ;  A  and  B  can  earn  it  in 
18  days ;  A  and  C  can  earn  it  in  21  days ;  in  what  time  can  each 
earn  it  alone  ? 

13.  A  and  B  can  earn  $40  in  6  days ;  A  and  C  can  earn  $54 
in  9  days ;  B  and  C  can  earn  $80  in  15  days :  what  can  each 
earn  per  day  ? 

1j^.  a  cistern  is  filled  by  8  pipes ;  the  first  and  second  fill  it 
in  1  hr.  10  min. ;  the  first  and  third  in  1  hr.  24  min. ;  and  the 
second  and  third  in  2  hr.  20  min. :  in  what  time  can  each  fill  it  ? 

15.  Find  three  numbers,  such  that  the  first  together  with  | 
of  the  second  is  equal  to  19 ;  ^  of  the  second  with  |  of  the  third 
is  equal  to  28 ;  and  \  of  the  third  with  ^  of  the  first  is  equal  to 
the  second. 

16.  A  certain  number  is  composed  of  three  digits  whose  sum 
is  15.     The  digit  in  units'  place  is  8  times  that  in  the  hundreds' 


ADVANCED   ARITHMETIC. 

place ;  and  if  896  be  added  to  the  number,  the  digits  will  be 
reversed  in  order.     Find  the  number. 

17.  A  jeweler  sold  three  rings.  The  price  of  the  first  with  -J 
the  sum  of  the  second  and  third  was  $25 ;  the  price  of  the  sec- 
ond with  i  of  the  sum  of  the  first  and  third  was  $26 ;  and  the 
price  of  the  third  with  \  the  sum  of  the  first  and  second  was 
$29.     What  was  the  price  of  each  ? 

18.  Find  four  numbers  such  that,  by  adding  each  to  twice 
the  sum  of  the  remaining  three,  you  will  obtain  46,  43,  41,  and 
88,  respectively. 

19.  A  gives  to  B  and  C  twice  as  much  money  as  each  of  them 
has ;  B  then  gives  to  A  and  C  twice  as  much  as  each  of  them 
has ;  and  C  then  gives  to  A  and  B  twice  as  much  as  each  of 
them  has ;  each  then  has  $27.     How  much  had  each  at  first  ? 

20.  A,  B,  C  and  D  together  have  $46.  After  B  gives  to  A  J 
of  his  money,  C  to  B  i  of  his,  and  D  to  C  J  of  his,  each  has 
$10.     How  much  had  each  at  first  ? 

D.     PROBLEMS  CONTAINING  QUADBATIC  EQUATIONS. 

179.  Quadratic  Equations. — An  equation  that  con- 
tains the  second  power  (or  square)  of  the  unknown  number  is 
called  a  Quadratic  Equation.  An  equation  that  contains 
no  form  of  the  unknown  number  except  the  square  is  called  an 
incomplete  quadratic  equation.  An  equation  that  contains  both 
the  first  and  second  powers  of  the  unknown  number  is  called  a 
complete  quadratic  equation.     Thus : 

(1)  a:2=64.     (Incomplete.) 

(2)  a:2+4a:  =  12.     (Complete.) 

When  an  equation  is  of  the  form  of  (1),  the  value  of  the  un- 
known number  may  be  found  by  extracting  the  square  root  of  the 
equation. 

When  an  equation  is  of  the  form  of  (2),  something  must  be 


394  ADVANCED    ARITHMETIC. 

added  to  render  the  first  member  a  perfect  square ;  then,  the 
value  of  the  unknown  number  may  be  found  by  extracting  the 
square  root  of  the  equation. 

Principle  :  Extracting  the  square  root  of  each  member  of  an 
equation  extracts  the  square  root  of  the  equation. 

EXAICPIiSS. 

1.  6x^=54.     Find  the  values  of  x. 

Process:  (1)  Qx^  =  5i. 
iof  (1H(2)  ^r2  =  9. 
t/(2)  =  (3)  x=  ±3,  result. 

Note.— +3x+3  =  9  and  -3x -3  =  9.  Then,.r  may  equal  +3  or  -3. 
When  a  number  is  true  for  either  sign,  the  double  sign  =h  is  used.  This 
sign  is  read  plus  or  minus,  and  should  be  used  whenever  we  extract  the 
square  root. 

2.  — -1 — =T7^H — .     Find  the  values  of  x. 
3     a;      12      x 


Process  : 

^^'  3+:r     12+a;' 

12:rx(l): 

=  (2)  4.r2+36  =  .r2+144. 

(3)  4.r2-.r2  =  144-36. 

(4)  3x2  =  108. 

(5)  a:2  =  36. 

-/(5)  = 

=  (6)  x=  zb6,  result. 

We  are  now  to  study  how  to  complete  the  square  of  the  first 
member  in  a  complete  quadratic  equation.     We  know  that — 

(a;-|-a)2=:a:2+2«a:+a2,  and 
(x—ay=x^—2ax+a^. 

How  could  we  get  the  third  term  in  these  results,  if  we  had 
the  second  term  ? 

Answer :  Take  half  of  the  2  a  and  square  it. 
How  can  we  get  the  third  term  in 

x^  +  12x  and 
x^-12x? 


A 


ADVANCED   ARITHMETIC.  395 

Answer  :  Take  half  of  the  12  and  square  it.     Thus  : 

a;2+12a;+36 

a;2-12a;+36. 
Rule  for  Completing  the  Square  :    Add  the  square  of  half 
the  coefficient  of  the  first  power  of  the  unknown  number, 
8.  x'^-\'2x-\Z.     Find  the  values  of  a;.  . 

Process:  (1)  a:2-12:c  =  13. 
Completing  the  sq.,  (2)  a:2-12a:+36  =  13  +  36. 

(3)  a:2- 12  0:4-36  =  49. 

V(3)  =  (4)  rr-6  =  ±7. 

(5)  a:  =  6±:7  =  13or  -1,  results. 

4.  5a;2+20a:=60.     Find  the  values  of  x. 

Note. — Always  divide  by  the  coefficient  of  x^  before  completing  the 
square. 

Process:  (1)  5a:2+20a:  =  60. 

(2)  cc^+ A x=l2. 

(3)  a;2+4.r+4  =  12+4  =  16. 
V73r=(4)  3:+2==h4. 

(5)  a:  =  2or  -6. 

5.  x^-\-l\  a:=:  — 18.     Find  the  values  of  x. 

Process:  (1)  x^-\-llx=  -18. 

(2)  a:2+lla'+i|i=  -l8+i|i=-<^ 

■i/(2)  =  (3)^+V  =  ±|. 

(4)  x=  -9  or  -1. 

x       44 

6.  -r  —  — ^=4.     Find  the  values  of  x. 
4     x  —  2 

Process:  (1) o"** 

(2)  a;2- 2  a; -176  =  16.^-32. 

(3)  a;2_2a;-16aj  =  176-32  =  144. 

(4)  a;2-18a;  =  144. 

(5)  a;2-18a;  +  81  =  144+81  =  225. 
-v'^=(6)  a- -9  =  ±15. 

a-  =  24or  -6. 


396  ADVANCED    ARITHMETIC. 

EXERCISE  CLIX. 

1.  What  is  a  quadratic  equation  ? 

2.  Define  incomplete  quadratic  equation.     Complete  quadratic 
equation. 

3.  Give  the  principle  governing  extracting  the  square  root  of 
an  equation. 

If..  How  do  you  complete  the  square  in  a  complete  quadratic 
equation  ? 

Find  the  values  of  x  in — 

5,  a;2_^2a;=8.  15.  3a:2+7  =  43+2a:2. 

8,  a;2-8a:=33.  17.  a:2+10a;  =  ll. 

9.  x'^-x=i.  18,  5a;2-40a;  =  165. 

10.  x'^-\-x=-\,  19.  4.T2-4a:=3. 

11.  a;2-14=-24.  20.  4a:2+20.r=-9. 

12.  a;2+5a;=-2i.  a:+2      4-a;_7 
i5.  a:2=:625.  a;-l        2a;  ~3' 

U.   ^-15. 


180.  Problems. 


EXAMPIiES. 


1.  A  bought  a  lot  of  flour  for  $126.  If  the  number  of  $'8 
per  barrel  was  equal  to  f  of  the  number  of  barrels  bought,  find 
the  number  of  barrels  and  the  price  per  barrel. 

Solution :  Let  a;  =  number  of  barrels. 
Then,  |a:  =  price  per  barrel, 
and  (1)  xxlx  =  12Q. 
(2)  1^2=126. 

(3)  .t2  =  441. 

V{S)   =(4)  .r=zb21. 

(5)  I  of  21  =  6. 
Answer,  21  bbls.  @  $6  eac|^ 


ADVANCED    ARITHMETIC.  397 

Note. — While  the  negative  answer  satisfies  the  equation  it  will  not 
satisfy  the  problem,  and  the  positive  answer  only  is  used  in  the  problem. 

2.  Divide  12  into  two  parts,  such  that  their  product  will  be 

11. 

Solution :  Let  x  =  one  part. 

Then,  12  -  a:  =  the  other  part 
and(l)  a;(12-:r)  =  ll. 

(2)  12:r-.r2  =  ll. 

(3)  :c2_i2x=-ll. 

(4)  x^-12x+SQ=  -11+36  =  25. 
(5)x-6  =  =b5. 

(6)  :c  =  ll,  or  1. 

(7)  12-a:  =  l,  orll. 
Answer,  1  and  11. 

3.  If  we  should  reduce  the  length  of  an  edge  of  a  cube  1  inch, 
the  volume  would  be  reduced  397  cu.  in.  What  is  the  volume 
of  the  cube  ? 

Note. — If  we  commence  with  the  cube  left  and  think  of 
building  the  1  inch  back  on  to  it,  let  .r  =  one  edge  of  that 
cube.    Then,  we  must  add  a  block  x  in.  1.,  x  in.  w.,  1  in. 

th.,  to  each  of  three  faces  of  the  cube .        =Sx^, 

a  rectangular  solid  :r  in.  1.,  1  in.  w.,  1  in.  th.,  to  each  of 

three  edges =Sx, 

and  a  1-inch  cube  in  the  corner =1 


The  whole  addition =3 a.2+3 x-^1 

Then,  (1)  3a,i+3a:+l  =  397. 

(2)  3x2  +  3  a  =397-1  =  396. 

(3)  x^+x  =  lS2. 

(4)a:2+:r+i=132+i  =  ^. 

V{i)  =  (b)  x+l  =  ±-%K 
(6)  ^  =  11,  -12. 
The  old  cube  11  in.  +  l  in.  =  12  in.,  answer. 


898  ADVANCED    ARITHMETIC. 

EXERCISE  CLX. 

1.  Find  two  numbers,  such  that  their  diflference  is  5  and  their 
product  is  66. 

2.  Find  that  number  whose  square  plus  6  times  the  number 
equals  55. 

3.  The  sum  of  two  numbers  is  12,  their  product  is  85.  Find 
the  numbers. 

Jf..  A  rectangular  field  is  twice  as  long  as  it  is  wide,  and  con- 
tains 20  acres.     How  wide  is  it  ? 

5.  The  product  of  two  consecutive  integers  exceeds  six  times 
their  sum  by  6.     Find  the  numbers. 

6.  If  we  cut  enough  from  a  cubical  block  to  make  its  dimen- 
sions 1  inch  shorter,  it  will  lose  1657  cu.  in.  Find  the  size  of 
the  block. 

7.  If  we  cut  enough  from  a  cubical  block  to  make  its  dimen- 
sions 2  inches  shorter,  it  will  lose  728  cu.  in.  Find  the  size  of 
the  block. 

8.  The  hypotenuse  of  a  right  triangle  is  50  inches,  its  base 
equals  one-half  of  its  perpendicular.  Find  the  base  and  per- 
pendicular. 

Hint:  Let  x  =  base. 

Then,  2. t  =  perpendicular, 
and  (1)  a;2+(2af  =  502,  or 
(2)  :i-2 +4^-2  =  2500. 

9.  Find  the  altitude  of  an  equilateral  triangle  if  one  side  is 
40  inches. 

Hint :  A  perpendicular  from  the  vertex  to  the  base  is  the 
altitude.  It  divides  the  triangle  into  two  equal  right  tri- 
angles. The  base  of  each  right  triangle  is  one-half  of  the  hy- 
potenuse (40  in.).    Letting  :r  =  the  altitude,  we  have  — 

402  =  202+.r2. 

10.  A  merchant  sold  a  piece  of  cloth  for  $24,  gaining  as  many 
%  as  the  number  of  dollars  the  cloth  cost.     Find  the  cost. 


ADVANCED    ARITHMETIC.  399 

11.  A  tree  90  ft.  high  was  blown  over  by  a  storm  so  that  the 
top  touched  the  ground  40  ft.  from  the  tree,  while  the  other 
end  of  the  part  broken  off  rested  on  the  stump.  How  much 
was  broken  off  ? 

12.  The  base  of  a  rectangle  is  11  ft.  longer  than  its  altitude. 
If  its  area  is  900  sq.  ft.,  find  the  length  of  each  side. 

13.  Find  the  altitude  and  area  of  a  triangle  whose  sides  are 
15,  12,  and  8. 

Note. — When  you  drop  the  altitude  on  the  base  15,  you  have  two 
right  triangles.  Let  a:  =  the  base  of  one  triangle  and  15 -a:  the  base  of 
the  other.    Then, 

(Altitude)2  =  122-(15-:j:)2  =  82-a:2.    (Why?) 
Find  the  value  of  x,  and  complete  the  solution. 

E.      PROGRESSIONS. 

181.  Aritlimetical  Progressions.  —  An  Arith- 
metical Progression  is  a  series  of  numbers  that  in- 
crease or  decrease  by  a  common  difference.     Thus, 

I,  4,  7,  10,  18, 

is  an  increasing  arithmetical  progression.     The  common  differ- 
ence is  3. 

II,  9,  7,  3,  1, 

is  a  decreasing  arithmetical  progression.     The  common  differ- 
ence is  —2. 

TERMS. 

a,  the  first  term. 

I,   the  last  term. 

n,  the  number  of  terms. 

(Z,  the  common  difference. 

S^  the  sum  of  the  series. 


400  ADVANCED   ARITHMETIC. 

Developing  the  formulas  for  arithmetical  progressions. 

I.  The  formula  for  the  last  term:  Using  the  terms  given  above,  we  may 
write  an  arithmetical  progression  or  series.    Thus, 

(1)        (2)  (3)  (4)  (5)         (n) 

S  =  a+{a-\-d)  +  {a+2d)  +  {a+3d)  +  {a-{-4d) (  ). 

Observe  (1)  that  each  term  of  the  series  has  a,  and  that  each  term 
after  the  first  has  d;  (2)  that  the  coeflRcient  of  d  is  always  1  less  than  the 
number  of  the  term.    Then,  for  the  nth  term,  the  coefficient  of  d  is  (n-l). 

From  these  data,  we  may  write  the  last  term.     Thus, 

l=a-{-{n—l)  d. 

Relation  :  In  an  arithmetical  progression,  the  last  term  is  equal 
to  the  first  term,  plus  the  common  difference  times  the  number  of 
terms  less  1. 

II.  The  formula  for  the  sum  of  the  series  :  An  arithmetical  series  may 
be  written  thus : 

(1)  S  =  a+{a+d)+{a+2d) +(Z-2d)  +  (Z-rf)  +  Z; 

also,  (2)  S  =  l  +  {l-d)  +  {l-2d) +(a+2rf)-f(a  +  d)  +  a. 

Adding,  (3)  2S  =  {a  +  l)  +  {a-\-l)  +  {a+l) +{a  +  l)  +  {a  +  l)+{a+l). 

Since  there  are  n  terms  and  each  has  (a+Z), 
(3)  =  (4)  2S  =  n{a-\-l);  or, 

S=l(a+l). 

Relation  :  The  sum  of  an  arithmetical  series  is  equal  to  the 
product  of  one-half  of  the  number  of  terms  and  the  sum  of  the  ex- 
tremes (the  first  and  last  terms). 

EXAMPLES. 

1.  Find  the  7th  term  of  the  series,  whose  1st  term  is  5,  and 
the  common  difference  is  9. 

Solution:  (1)  l  =  a-\-{n-l)d. 

(2)  Z  =  5+(7-l)9  =  59,  answer. 

2.  Find  the  10th  term  of  the  series  2,  6,  10,  etc. 

Solution:  (1)  a  =  2.     (Why?) 

(2)  d  =  4.     (Why?) 

(3)  Z  =  2+(10-l)4  =  38,  answer. 


J 


ADVANCED    ARITHMETIC.  401 

3.  Find  the  sum  of  a  series  of  19  terms,  whose  first  term  is  6 
and  whose  last  term  is  96. 

Solution:  (1)  S  =  ^{a  +  l). 

iq 
(2)  ^=^(6+96)  =  969,  answer. 

4..  Find  the  last  term  and  the  sum  of  the  series,  5,  5,  etc., 

to  25  terms. 

Solution:  (1)  a  =  5. 

(2)  d  =  3. 

(3)  n  =  25. 

(4)  ?  =  5+(25-l)3  =  77,  1st  answer. 

(5)  s  =  V-  (5 +77)  =  1025,  2d  answer. 

5.  Find  the  last  term  and  the  sum  of  the  series  12^  9^  6,  etc., 

to  15  terms. 

Solution  :   (1)  a  =  12. 

(2)rf=-3.    (Why?) 

(3)  n  =  15. 

(4)  Z  =  12-f(15-l)(-3). 

(5)  ?  =  12-42=  -30,  1st  answer. 

(6)  s  =  y  [12  +  (-30)]. 

(7)  s  =  JgS  X  - 18  =  - 135,  2d  answer. 

Note. — Be  careful  to  handle  the  signs  correctly,  and  a  minus  common 
diflference  will  give  you  no  trouble. 

6.  Find  the  first  term  of  a  series  whose  last  term,  common 
difference,  and  number  of  terms  are  respectively  27,  2,  and  11. 

Solution:  (1)  27  =  a+(ll-l)  2  =  a+20. 
Trans.,  (2)  a  =  27 -20  =  7,  answer. 

7.  Find  the  number  of  terms  of  the  series  whose  first  term, 
last  term,  and  sum  are  respectively  —1,  44,  215. 

Solution:   (1)  215  =  -|^(-l+44)  =  -|-x43. 

(2)  ^4^  =  215. 

,„,         215x2    ,_ 

(3)  n  =  —^ —  =  10,  answer. 


402 


ADVANCED    ARITHMETIC. 


EXERCISE  OLXT. 

1.  What  is  an  arithmetical  i)rogression  or  series  ? 

2,  Name  and  explain  the  terms  used. 

S.  Write  a  series  of  8  terms  whose  first  term  is  5  and  whose 
common  difference  is  12. 

Jf..  Write  a  series  of  12  terms,  whose  first  term  is  12  and 
whose  common  difference  is  —5. 

5.  Develop  the  formula  for  finding  I.     Give  the  relation. 

6.  Develop  the  formula  for  finding  s.     Give  the  relation. 


No. 

I 

a 

n 

d 

s 

7. 

9 

7 

9 

5 

9 

8. 

? 

-5 

12 

-3 

? 

9. 

? 

27 

18 

-4 

? 

10. 

2 

? 

7 

-1 

? 

11. 

24 

-15 

14 

? 

? 

12. 

116 

1 

? 

5 

? 

13. 

81 

? 

'7 

7 

? 

n. 

m 

1 

27 

? 

? 

15. 

4 

? 

f 

? 

16. 

24 

8 

? 

? 

40 

17. 

? 

27 

9 

? 

68 

18. 

H 

? 

12 

? 

-99 

19. 

15i 

? 

? 

*l 

168 

20. 

9 

? 

17 

4 

-891 

Note.— The  last  two  (Nos.  19  and  20)  will  require  two  equations.  Sub- 
stitute in  both  formulas.  This  will  give  you  two  equations  to  solve  fcr 
two  unknown  numbers. 


21.  40  potatoes  are  2  yd.  apart  and  the  first  is  2  yd.  from  a 
basket.  How  far  will  a  boy  travel,  who  gathers  them  and  puts 
them  into  the  basket  one  at  a  time  ? 


ADVANCED    ARITHMETIC.  403 

182.  Falling'  Bodies. — Falling  of  bodies  furnishes  an 
example  of  arithmetical  series.  A  body,  if  left  unsupported, 
will  fall  by  its  own  weight  during  the  1st  second,  16.08  ft.  (N. 
Y.);  during  the  2d  second,  48.24  ft.;  during  the  8d  second, 
80.4  ft. ;  and  so  on.     Thus, 


16.08,  48.24,  80.4,  112.56, 


Note. —  Observe  that  in  this  series  the  first  term  {a)  is  16.08,  and  the 
common  difference  (d)  is  32.16. 

EXAMPIiES. 

1.  How  far  will  a  body  fall  during  the  10th  second  ? 

Solution:  (1)  a  =  16.08. 

(2)  rf  =  32.16. 

(3)  n  =  10. 

(4)  l  =  a  +  {n-l)d. 

(5)  ?  =  16.08+9x32.16  =  305.52. 
Answer,  305.52  ft. 

2.  How  far  will  a  body  fall  in  8  seconds  ? 

Solution:  (!)  Z  =  16.08 +7x32.16  =  241.2. 
(2)  ^  =  1(16.08  +  241.2)  =  1029.12. 
Answer,  1029.12  ft. 

3.  A  body  is  thrown  downward  so  that  it  travels  70  ft.  the 
first  second.     How  far  does  it  fall  in  5  seconds  ? 

Note. —  Here  a  =  70. 

Solution:  (1)  Z  =  70+4x32.16  =  198.64. 
(2)  >S  =  1(70+198.64)  =  671.6. 
Answer,  671.6  ft. 

In  the  falling  of  bodies  it  has  been  determined  that  the  ve- 
locity increases  82.16  ft.  per  second.  This  is  the  acceleration. 
Using  82.16  for  a  in  formula  on  page  809,  we  have — 

F=32.16^ 


404  ADVANCED   ARITHMETIC. 

Relation  :  Abstractly,  the  velocity  of  a  falling  body  is  32.16 
times  the  time  expressed  in  seconds. 

If.  A  body  strikes  the  ground  with  a  velocity  of  160.8  ft.  per 
second.  How  many  seconds  has  it  been  falling  and  how  far 
has  it  fallen  ? 

Solution:  (1)  F=32.16f. 

(2)  160.8  =  32.16  <. 

160.8 
^^^  ^  =  32:i6  =  ^- 

Time,  5  seconds. 

(5)  Z  =  16.08+4x32.16  =  144.72. 

(6)  «  =  1(16.08  +  144.72)  =  402. 
Distance,  402  ft. 

Note. — A  body  thrown  upward  will  decrease  in  its  travel  just  as  a 
falling  body  increases  in  its  travel. 

5.  How  high  will  a  body  rise,  if  thrown  upward  with  a  veloc- 
ity of  321.6  ft.  per  second  ? 

Solution:  (1)  F=  32.16 i. 

(2)  321.6  =  32.16  <. 

Time  in  rising,  10  sec. 

Note. — If  it  was  10  sec.  rising,  the  last  second  it  rose  16.08  ft.  To  find 
how  far  it  rose  the  first  second,  consider  from  the  top  downward  and  get 
the  10th  term  of  the  series. 

(4)  Z  =  16.08+9x32.16  =  305.52. 

(5)  ;S'  =  J^(16.08+305.52)  =  1608. 
Distance,  1608  ft. 

EXERCISE  CLXII. 

1.  How  far  will  a  body  fall  in  1  sec?     In  2  sec?     In  8  sec? 

2.  Do  the  distances  fallen  during  the  successive  seconds 
form  an  arithmetical  progression  ?  If  so,  what  is  the  common 
difference  ? 

3.  What  is  the  formula  for  the  velocity  of  falling  bodies  ? 


ADVANCED    ARITHMETIC.  405 

Jf.  A  body  thrown  upward  is  gone  6  sec.  How  much  time  is 
taken  in  rising  ?     How  much  in  falling  ? 

5.  How  fast  will  a  body  be  falling  at  the  end  of  the  6th  sec- 
ond of  its  descent  ?     Give  its  fall  during  the  7th  second. 

6.  A  boy  throws  a  stone  over  a  tree ;  it  strikes  the  ground  in 
4  sec.     How  high  is  the  tree  ? 

7.  A  body  strikes  the  ground  with  a  velocity  of  96.48  ft.  per 
sec.     How  long  has  it  been  falling  ? 

8.  A  body  is  thrown  downward  with  a  velocity  of  60  ft.  per 
second.  What  will  its  velocity  be  in  6  sec?  How  far  will  it 
fall?     (a =60+ 16.08.     Why?) 

9.  A  tower  is  321.6  feet  high.  With  what  velocity  will  a 
stone,  let  fall  from  the  top,  strike  the  ground  ? 

Plan  :  (1)  Find  the  time  and  (2)  the  velocity. 

183.  Geometrical  Progressions.— A  Geometri- 
cal Progression  is  a  series  of  numbers  that  increase  or 
decrease  by  a  common  ratio. 

1,  4,  16,  64,  etc., 
is  an  increasing  geometrical  series.     The  common  ratio  is  4. 

24,  12,  6,  3,  etc., 
is  a  decreasing  geometrical  series.     The  common  ratio  is  -J. 

TERMS. 

a,  the  first  term. 

ly   the  last  term. 

r,  the  common  ratio. 

n,  the  number  of  terms. 

S,  the  sum  of  the  series. 


406  ADVANCED    ARITHMETIC. 


Developing  the  formulas  for  geometrical  series. 

I.  The  formula  for  the  last  term  :  Using  the  terms  given  above,  we  may 

•ite  a  geometrical  series. 


(1)     (2)       (3)        (4) (n). 

S  =  a+ar+ar^+ar^ (  ). 

Observe  (1)  that  each  term  of  the  series  has  a  factor  a,  and  (2)  every 
term  after  the  first  has  a  factor  r  whose  exponent  is  1  less  than  the  number 
of  the  term.    Then,  we  may  write  the  last  term  — 

l=ar^-'i^ 

Note. — n  -1  is  the  exponent  of  r. 

Relation:  In  a  geometrical  series,  the  last  term  is  equal  to  the 
first  term  times  the  ratio  raised  to  a  power  whose  exponent  is  1  less 
than  the  number  of  terms. 

II.  The  formula  for  the  sum  of  the  series  :  The  series  may  be  written  in 
this  form : 

(1)     (2)       (3)        (4) (n) 

(1)  S  =  a-\-ar-\-a'i^+ar^ I. 

(1)       (2)        (3) {n-D     (n) 

rx(l)  =  (2)  >Sr  =  ar+ar2+ar3 I    +Zr 

(2)-(l)  =  (3)  Sr-S  =  lr-a. 

(4)  S{r-l)  =  lr-a',or, 

r  —  1 

Note. — When  (1)  is  subtracted  from  (2),  all  the  terms  in  the  second 
members  disappear,  except  the  first  term  in  (1)  and  the  last  term  in  (2). 

Relation  :  The  sum  of  a  geometrical  series  is  equal  to  the  frac- 
tion whose  numerator  is  the  last  term  times  the  ratio ^  minus  the 
first  term,  and  lohose  denominator  is  the  ratio  less  1. 

EXAMPLES. 

1.  Find  the  fifth  term  of  the  series  whose  first  term  and  com- 
mon ratio  are  respectively  4  and  3. 

Solution:  (1)  l  =  ar'^~'^ 


(2)  Z  =  4x  3^  =  324,  answer. 


I 


ADVANCED   ARITHMETIC.  407 

2.  Find  the  ratio  of  the  series  whose  first  term,  number  of 
terms  and  last  term  are  respectively  20,  4,  and  540. 

Solution:  (1)  l  =  ar'^~'^ 

(2)  540  =  20r3. 

(3)  1^  =  21. 

(4)  r=  f27  =  3,  answer. 

3.  Find  the  sum  of  8  terms  of  the  series,  3,  6,  12,  etc. 

Solution:  (1)  r  =  2. 

(2)  Z  =  3x27  =  384. 

(3)^=^^«. 

(4)  S  =  ^""^^^"^  =  765,  answer. 

4.  Find  the  sum  of  the  infinite  series, 

1+i+i+i+etc. 

Note. — An  infinite  series  continues  forever.  This  series  is  decreasing 
and  its  last  term  approaches  infinitely  near  to  0 — so  near  that  no  other 
value  represents  it  so  accurately  as  0.  Then,  we  have  a  =  l,  r  =  i,  and 
1  =  0. 

Solution :  S  =  ^ — :r  =  — r  =  2,  answer. 

5.  Find  the  number  of  terms  in  the  series  whose  1st  term, 
last  term,  and  ratio  are  respectively  5,  1280,  4. 

Solution:  (1)  1280  =  5x4"—^ 

(2)  4""-^  =  256. 

(3)  4^  =  256. 

(4)  n-l  =  4. 

(5)  n  =  5,  answer. 

NoTE.-^To  obtain  (3),  you  know  that  4  raised  to  some  power  equals  256. 
Proceed  by  expanding  4,  and  when  you  reach  the  4th  power  you  will 
have  256. 

EXEECISE  CLXIII. 

1.  What  is  a  geometrical  progression  or  series  ? 

2.  Give  and  explain  all  terms  used  in  geometrical  series. 


408 


ADVANCED    ARITHMETIC. 


3.  Write  a  geometrical  series  of  4  terms,  whose  1st  term  is  5 
and  whose  ratio  is  7. 

J^..  Write  a  series  of  4  terms,  whose  1st  term  is  4  and  whose 
ratio  is  ^. 

5.  Write  5  terms  of  the  series  whose  1st  term  is  4  and  whose 
ratio  is  —3. 

6.  Develop  the  formula  for  I.     Give  the  relation. 

7.  Develop  the  formula  for  S.     Give  the  relation. 

8.  What  is  an  infinite  series  ?  If  the  infinite  series  is  de- 
creasing, what  does  the  I  approach  ? 


No. 

I 

a 

n 

r 

»S' 

9. 

? 

4 

7 

5 

? 

10. 

? 

-3 

10 

3 

9 

11.' 

? 

1 

8 

-4 

9 

12. 

2 

infinite 

f 

? 

13. 

64 

1 

? 

2 

? 

n- 

? 

4 

? 

3 

118096 

15. 

13122 

? 

9 

3 

? 

16. 

? 

2 

8 

i 

? 

17. 

375000 

? 

7 

5 

468769 

18. 

? 

infinite 

^ 

n 

i 


YC  49538 


lujuei 


